Math Ncert Exemplar 2019 Solutions for Class 11 Science Maths Chapter 3 - Trigonometric Functions

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Page No 52:

Question 1:

Prove that $\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \frac{1 + \sin A}{\cos A}$

Answer:

To prove: $\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \frac{1 + \sin A}{\cos A}$
LHS = $\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1}$
$= \frac{\tan A + \sec A - (s e c^{2} x - \tan^{2} x)}{\tan A - \sec A + 1}$
$= \frac{\tan A + \sec A - \{(secx + tanx) (secx - tanx)\}}{\tan A - \sec A + 1} = \frac{(\tan A + \sec A) (1 - s e c x + \tan x)}{\tan A - \sec A + 1}$
$= \tan x + s e c x = \frac{\sin x}{\cos x} + \frac{1}{\cos x}$
$= \frac{1 + \sin x}{\cos x}$ = RHS
Hence proved.

Page No 52:

Question 2:

If $\frac{2 \sin α}{1 + \cos α + \sin α} = y$ , then prove that $\frac{1 - \cos α + \sin α}{1 + \sin α}$ is also equal to y.

Answer:

Given: $y = \frac{2 \sin α}{1 + \cos α + \sin α}$
$= \frac{2 \sin α}{1 + \cos α + \sin α} \times \frac{1 + \sin α - \cos α}{1 + \sin α - \cos α} = \frac{2 \sin α (1 + \sin α - \cos α)}{{(1 + \sin α)}^{2} - \cos^{2} α} = \frac{2 \sin α (1 + \sin α - \cos α)}{1 + 2 \sin α + \sin^{2} α - \cos^{2} α}$
$= \frac{2 \sin α (1 + \sin α - \cos α)}{2 \sin α + 2 \sin^{2} α} = \frac{2 \sin α (1 + \sin α - \cos α)}{2 \sin α (1 + \sin α)} = \frac{1 + \sin α - \cos α}{1 + \sin α}$
Hence proved.

Page No 52:

Question 3:

If m sin θ = n sin (θ + 2α), then prove that tan (θ + α) $\cot α = \frac{m + n}{m - n}$

Answer:

Given: m sin θ = n sin (θ + 2α)
$\Rightarrow \frac{m}{n} = \frac{\sin (θ + 2 α)}{\sin θ}$
Applying componendo and dividendo,
$\Rightarrow \frac{m + n}{m - n} = \frac{\sin (θ + 2 α) + \sin θ}{\sin (θ + 2 α) - \sin θ} = \frac{2 \sin \frac{2 θ + 2 α}{2} \cos \frac{2 α}{2}}{2 \cos \frac{2 θ + 2 α}{2} \sin \frac{2 α}{2}} = \tan (θ + α) c o t θ$
$\Rightarrow \tan (θ + α) c o t θ = \frac{m + n}{m - n}$
Hence proved.

Page No 52:

Question 4:

If $\cos (α + β) = \frac{4}{5} and \sin (α - β) = \frac{5}{13}$ , where α lie between 0 and $\frac{π}{4}$ , find the value of tan 2α

Answer:

Given: $\cos (α + β) = \frac{4}{5} and \sin (α - β) = \frac{5}{13}$
Therefore, $\sin (α + β) = \frac{3}{5} and \cos (α - β) = \frac{12}{13}$
Now, $2 \sin α \cos β = \sin (α + β) + \sin (α - β) = \frac{3}{5} + \frac{5}{13} = \frac{64}{65} . . . . . (1)$
and $2 \cos α \cos β = \cos (α + β) + \cos (α - β) = \frac{4}{5} + \frac{12}{13} = \frac{112}{65} . . . . . (2)$
Dividing (1) by (2), we have $\tan α = \frac{\frac{64}{65}}{\frac{112}{65}} = \frac{4}{7}$
So, $\tan 2 α = \frac{2 \tan α}{1 - \tan^{2} α} = \frac{2 \times \frac{4}{7}}{1 - \frac{16}{49}} = \frac{8}{7} \times \frac{49}{33} = \frac{56}{33}$ .

Page No 53:

Question 5:

If $\tan x = \frac{b}{a}$ , then find the value of $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}$

Answer:

Given: $\tan x = \frac{b}{a}$
Applying componendo and devidendo, we get $\frac{a + b}{a - b} = \frac{1 + \tan x}{1 - \tan x} and \frac{a - b}{a + b} = \frac{1 - \tan x}{1 + \tan x}$
Therefore, $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}$
$= \sqrt{\frac{1 + \tan x}{1 - \tan x}} + \sqrt{\frac{1 - \tan x}{1 + \tan x}} = \frac{(1 + \tan x) + (1 - \tan x)}{\sqrt{1 - \tan^{2} x}} = \frac{2}{\sqrt{1 - \tan^{2} x}}$
$= \frac{2}{\sqrt{1 - \frac{\sin^{2} x}{\cos^{2} x}}} = \frac{2 \cos x}{\sqrt{\cos^{2} x - \sin^{2} x}} = \frac{2 \cos x}{\sqrt{\cos 2 x}}$
Hence, $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}$ $= \frac{2 \cos x}{\sqrt{\cos 2 x}}$ .

Page No 53:

Question 6:

Prove that $\cos θ \cos \frac{θ}{2} - \cos 3 θ \cos \frac{9 θ}{2} = \sin 7 θ \sin 8 θ$ .

Answer:

To prove: $\cos θ \cos \frac{θ}{2} - \cos 3 θ \cos \frac{9 θ}{2} = \sin 7 θ \sin 8 θ$
LHS $= \cos θ \cos \frac{θ}{2} - \cos 3 θ \cos \frac{9 θ}{2}$
$= \frac{1}{2} \{2 \cos θ \cos \frac{θ}{2} - 2 \cos 3 θ \cos \frac{9 θ}{2}\} = \frac{1}{2} \{\cos \frac{3 θ}{2} + \cos \frac{θ}{2} - \cos \frac{15 θ}{2} - \cos \frac{3 θ}{2}\}$
$= \frac{1}{2} (\cos \frac{θ}{2} - \cos \frac{15 θ}{2}) = \frac{1}{2} (2 \sin \frac{8 θ}{2} \sin \frac{7 θ}{2})$
$= \sin 4 θ \sin \frac{7 θ}{2}$ = RHS.

Page No 53:

Question 7:

If a cos θ + b sin θ = m and a sin θ – b cos θ = n, then show that a² + b² = m² + n²

Answer:

Given: a cos θ + b sin θ = m.....(1)
and a sin θ – b cos θ = n.....(2)
Squaring and adding (1) and (2), we have
LHS = $m^{2} + n^{2}$
$= {(a \cos θ + b \sin θ)}^{2} + {(a \sin θ - b \cos θ)}^{2} = a^{2} \cos^{2} θ + b^{2} \sin^{2} θ + 2 a b \sin θ \cos θ + a^{2} \sin^{2} θ + b^{2} \cos^{2} θ - 2 a b \sin θ \cos θ = a^{2} \cos^{2} θ + b^{2} \sin^{2} θ + a^{2} \sin^{2} θ + b^{2} \cos^{2} θ = a^{2} (\sin^{2} θ + \cos^{2} θ) + b^{2} (\sin^{2} θ + \cos^{2} θ)$
$= a^{2} + b^{2}$ = RHS.

Page No 53:

Question 8:

Find the value of tan 22°30'.

Answer:

To find: tan 22°30' = $\tan 22 \frac{1}{2} °$
Applying the formula $\tan 2 x = \frac{2 \tan x}{1 - \tan^{2} x}$ and putting $x = 22 \frac{1}{2} °$ ,
$\tan 45 ° = \tan 2 (22 \frac{1}{2}) ° = \frac{2 \tan 22 \frac{1}{2} °}{1 - \tan^{2} 22 \frac{1}{2} °} \Rightarrow 1 = \frac{2 \tan 22 \frac{1}{2} °}{1 - \tan^{2} 22 \frac{1}{2} °}$
Let $\tan 22 \frac{1}{2} ° = x$
So, $1 = \frac{2 x}{1 - x^{2}}$
$\Rightarrow 1 - x^{2} = 2 x \Rightarrow x^{2} + 2 x - 1 = 0$
$∴ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- 2 \pm \sqrt{4 + 4}}{2} = - 1 \pm \sqrt{2}$
Since the given angle is in the first quadrant so $x = \tan 22 \frac{1}{2} °$ must be positive.
Hence, $x = \tan 22 \frac{1}{2} ° = \sqrt{2} - 1$ .

Page No 53:

Question 9:

Prove that sin 4A = 4sinA cos³A – 4 cosA sin³A.

Answer:

To prove: sin 4A = 4sinA cos³A – 4 cosA sin³A
LHS = $\sin 4 A = \sin (3 A + A)$
$= \sin 3 A \cos A + \cos 3 A \sin A = (3 \sin A - 4 \sin^{3} A) \cos A + (4 \cos^{3} A - 3 \cos A) \sin A = 3 \sin A \cos A - 4 \cos A \sin^{3} A + 4 \sin A \cos^{3} A - 3 \cos A \sin A$
$= 4 \sin A \cos^{3} A - 4 \cos A \sin^{3} A$ = RHS

Page No 53:

Question 10:

If tanθ + sinθ = m and tanθ – sinθ = n, then prove that m² – n² = 4sinθ tanθ

Answer:

Given: tanθ + sinθ = m and tanθ – sinθ = n
LHS = $m^{2} - n^{2}$
$= {(\tan θ + \sin θ)}^{2} - {(\tan θ - \sin θ)}^{2} = \tan^{2} θ + \sin^{2} θ + 2 \tan θ \sin θ - \tan^{2} θ - \sin^{2} θ + 2 \tan θ \sin θ$
$= 4 \sin θ \tan θ$ = RHS

Page No 53:

Question 11:

If tan (A + B) = p, tan (A – B) = q, then show that tan 2 A = $\frac{p + q}{1 - p q}$

Answer:

Given: tan (A + B) = p and tan (A – B) = q
LHS = $\tan 2 A = \tan \{(A + B) + (A - B)\}$
$= \frac{\tan (A + B) + \tan (A - B)}{1 - \tan (A + B) \tan (A - B)}$
$= \frac{p + q}{1 - p q}$ = RHS

Page No 53:

Question 12:

If cosα + cosβ = 0 = sinα + sinβ, then prove that cos 2α + cos 2β = – 2cos (α + β).

Answer:

Given: cosα + cosβ = 0 = sinα + sinβ
Therefore, ${(\cos α + \cos β)}^{2} - {(\sin α + \sin β)}^{2} = 0$
$\Rightarrow \cos^{2} α + \cos^{2} β + 2 \cos α \cos β - \sin^{2} α - \sin^{2} β - 2 \sin α \sin β = 0 \Rightarrow (\cos^{2} α - \sin^{2} α) + (\cos^{2} β - \sin^{2} β) + 2 (\cos α \cos β - \sin α \sin β) = 0 \Rightarrow \cos 2 α + \cos 2 β + 2 \cos (α + β) = 0 \Rightarrow \cos 2 α + \cos 2 β = - 2 \cos (α + β)$
Hence proved.

Page No 53:

Question 13:

If $\frac{\sin (x + y)}{\sin (x - y)} = \frac{a + b}{a - b}$ , then show that $\frac{\tan x}{\tan y} = \frac{a}{b}$ .

Answer:

Given: $\frac{\sin (x + y)}{\sin (x - y)} = \frac{a + b}{a - b}$
$\Rightarrow \frac{a + b}{a - b} = \frac{\sin (x + y)}{\sin (x - y)}$
Applying componendo and dividendo,
$\Rightarrow \frac{(a + b) + (a - b)}{(a + b) - (a - b)} = \frac{\sin (x + y) + \sin (x - y)}{\sin (x + y) - \sin (x - y)} \Rightarrow \frac{2 a}{2 b} = \frac{2 \sin x \cos y}{2 \cos x \sin y} \Rightarrow \frac{a}{b} = \frac{\tan x}{\tan y} \Rightarrow \frac{\tan x}{\tan y} = \frac{a}{b}$
Hence, proved.

Page No 53:

Question 14:

If $\tan θ = \frac{\sin α - \cos α}{\sin α + \cos α}$ , then show that $\sin α + \cos α = \sqrt{2} \cos θ$ .

Answer:

Given: $\tan θ = \frac{\sin α - \cos α}{\sin α + \cos α}$
Dividing numerator and denominator on the right side by $\cos α$
$\Rightarrow \tan θ = \frac{\frac{\sin α - \cos α}{\cos α}}{\frac{\sin α + \cos α}{\cos α}} \Rightarrow \tan θ = \frac{\tan α - 1}{1 + \tan α} \Rightarrow \tan θ = \frac{\tan α - \tan \frac{π}{4}}{1 + \tan \frac{π}{4} \tan α} \Rightarrow \tan θ = \tan (α - \frac{π}{4})$
$\Rightarrow θ = α - \frac{π}{4} \Rightarrow \cos θ = \cos (α - \frac{π}{4}) \Rightarrow \cos θ = \cos α \cos \frac{π}{4} + \sin α \sin \frac{π}{4} \Rightarrow \cos θ = \cos \frac{π}{4} (\cos α + \sin α) \Rightarrow \cos θ = \frac{1}{\sqrt{2}} (\cos α + \sin α) \Rightarrow \sqrt{2} \cos θ = \cos α + \sin α$
Hence proved.

Page No 53:

Question 15:

If sinθ + cosθ = 1, then find the general value of θ.

Answer:

Given: $\sin θ + \cos θ = 1$
Dividing both sides by $\sqrt{1^{2} + 1^{2}} = \sqrt{2}$
$\Rightarrow \frac{1}{\sqrt{2}} \sin θ + \frac{1}{\sqrt{2}} \cos θ = \frac{1}{\sqrt{2}} \Rightarrow \sin \frac{π}{4} \sin θ + \cos \frac{π}{4} \cos θ = \frac{1}{\sqrt{2}} \Rightarrow \cos (θ - \frac{π}{4}) = \cos \frac{π}{4} \Rightarrow θ - \frac{π}{4} = 2 n π \pm \frac{π}{4}, (n \in Z)$
$\Rightarrow θ = 2 n π \pm \frac{π}{4} + \frac{π}{4}, (n \in Z) ∴ θ = 2 n π, 2 n π + \frac{π}{2}, (n \in Z)$

Page No 53:

Question 16:

Find the most general value of θ satisfying the equation tanθ = –1 and $\cos θ = \frac{1}{\sqrt{2}}$ .

Answer:

Given: tanθ = –1 and $\cos θ = \frac{1}{\sqrt{2}}$
Since cosθ is positive and tanθ is negative that means θ lies in 4th quadrant. Solving the two one by one:
$\tan θ = \tan (- \frac{π}{4}) \Rightarrow \tan θ = \tan (2 π - \frac{π}{4}) \Rightarrow \tan θ = \tan \frac{7 π}{4} ∴ θ = n π + \frac{7 π}{4}, (n \in Z)$
Also, $\cos θ = \frac{1}{\sqrt{2}}$
$\Rightarrow \cos θ = \cos (2 π - \frac{π}{4}) \Rightarrow \cos θ = \cos \frac{7 π}{4} ∴ θ = 2 n π \pm \frac{7 π}{4}, (n \in Z)$
Clearly, the most general solution which is common in both of the above solutions is $2 n π + \frac{7 π}{4}, (n \in Z)$ .

Page No 54:

Question 17:

If cotθ + tanθ = 2 cosecθ, then find the general value of θ.

Answer:

Given: cotθ + tanθ = 2cosecθ
$\Rightarrow \frac{\cos θ}{\sin θ} + \frac{\sin θ}{\cos θ} = \frac{2}{\sin θ} \Rightarrow \frac{\sin^{2} θ + \cos^{2} θ}{\sin θ \cos θ} = \frac{2}{\sin θ} \Rightarrow \frac{1}{\sin θ \cos θ} = \frac{2}{\sin θ} \Rightarrow 2 \sin θ \cos θ = \sin θ \Rightarrow 2 \sin θ \cos θ - \sin θ = 0 \Rightarrow \sin θ (2 \cos θ - 1) = 0 \Rightarrow s o n θ = 0, \cos θ = \frac{1}{2}$
$\Rightarrow = n π, \cos θ θ = \cos \frac{π}{3} \Rightarrow θ = n π, 2 n π \pm \frac{π}{3}, (n \in Z)$

Page No 54:

Question 18:

If 2sin²θ = 3cosθ, where 0 ≤ θ ≤ 2π, then find the value of θ.

Answer:

Given: 2sin²θ = 3cosθ
$\Rightarrow 2 (1 - \cos^{2} θ) = 3 \cos θ \Rightarrow 2 - 2 \cos^{2} θ = 3 \cos θ \Rightarrow 2 \cos^{2} θ + 3 c o s θ - 2 = 0 \Rightarrow 2 \cos^{2} θ + 4 c o s θ - c o s θ - 2 = 0 \Rightarrow 2 c o s θ (c o s θ + 2) - 1 (c o s θ + 2) = 0 \Rightarrow (2 c o s θ - 1) (c o s θ + 2) = 0 ∴ c o s θ = \frac{1}{2}, - 2$
Since $- 1 \leq c o s θ \leq 1$ so -2 is invalid.
Therefore, $c o s θ = \frac{1}{2} \Rightarrow c o s θ = c o s \frac{π}{3}, \cos (2 π - \frac{π}{3})$
$\Rightarrow θ = \frac{π}{3}, \frac{5 π}{3}$

Page No 54:

Question 19:

If secx cos5x + 1 = 0, where 0 < x ≤ $\frac{π}{2}$ , then find the value of x.

Answer:

Given: $s e c x \cos 5 x + 1 = 0$ , where $0 < x \leq \frac{π}{2}$
$\Rightarrow \frac{\cos 5 x}{\cos x} + 1 = 0 \Rightarrow \cos 5 x + \cos x = 0$
$\Rightarrow 2 \cos 3 x \cos 2 x = 0$ $(\cos x + \cos y = 2 \cos \frac{x + y}{2} \cos \frac{x - y}{2})$
$Either \cos 3 x = 0 or \cos 2 x = 0$
$Either 3 x = (2 n + 1) \frac{π}{2} or 2 x = (2 n + 1) \frac{π}{2} ∴ x = \frac{π}{6}, \frac{π}{2}, \frac{π}{4}$

Page No 54:

Question 20:

If sin (θ + α) = a and sin (θ + β) = b, then prove that cos 2(α – β) – 4ab cos (α – β) = 1 – 2a² – 2b²

Answer:

Given: sin (θ + α) = a and sin (θ + β) = b
Therefore, $\cos (θ + α) = \sqrt{1 - a^{2}}$ and $\cos (θ + β) = \sqrt{1 - b^{2}}$
$\cos (α - β) = \cos \{(θ + α) - (θ + β)\} = \cos (θ + α) \cos (θ + β) + \sin (θ + α) \sin (θ + β) = \sqrt{(1 - a^{2}) (1 - b^{2})} + a b$
And $\cos 2 (α - β) = 2 \cos^{2} (α - β) - 1$
$= 2 {[\sqrt{(1 - a^{2}) (1 - b^{2})} + a b]}^{2} - 1 = 2 [1 - a^{2} - b^{2} + a^{2} b^{2} + 2 a b \sqrt{(1 - a^{2}) (1 - b^{2})}] - 1$
Now, LHS = cos 2(α – β) – 4ab cos (α – β)
$= 2 - 2 a^{2} - 2 b^{2} + 2 a^{2} b^{2} + 4 a b \sqrt{(1 - a^{2}) (1 - b^{2})} - 1 - 4 a b \sqrt{(1 - a^{2}) (1 - b^{2})} - 4 a^{2} b^{2}$
= $= 1 - 2 a^{2} - 2 b^{2}$ = RHS
Hence proved.

Page No 54:

Question 21:

If cos (θ + Ï•) = m cos (θ – Ï•), then prove that $\tan θ = \frac{1 - m}{1 + m} \cot ϕ$ .

Answer:

Given: $\cos (θ + ϕ) = m \cos (θ - ϕ)$
$\Rightarrow \frac{\cos (θ + ϕ)}{\cos (θ - ϕ)} = m \Rightarrow \frac{\cos θ \cos ϕ - \sin θ \sin ϕ}{\cos θ \cos ϕ + \sin θ \sin ϕ} = m$
Dividing numerator and denominator by $\cos θ \sin ϕ$ on the left side,
$\Rightarrow \frac{c o t ϕ - \tan θ}{c o t ϕ + \tan θ} = m \Rightarrow \frac{c o t ϕ + \tan θ}{c o t ϕ - \tan θ} = \frac{1}{m}$
Applying componendo and dividendo,
$\Rightarrow \frac{2 c o t ϕ}{2 \tan θ} = \frac{1 + m}{1 - m} \Rightarrow \frac{c o t ϕ}{\tan θ} = \frac{1 + m}{1 - m} \Rightarrow \frac{\tan θ}{c o t ϕ} = \frac{1 - m}{1 + m} \Rightarrow \tan θ = \frac{1 - m}{1 + m} c o t ϕ$
â€‹Hence proved.

Page No 54:

Question 22:

Find the value of the expression $3 [\sin^{4} (\frac{3 π}{2} - α) + \sin^{4} (3 π + α)] - 2 \{\sin^{6} (\frac{π}{2} + α) + \sin^{6} (5 π - α)]$

Answer:

Given: $3 [\sin^{4} (\frac{3 π}{2} - α) + \sin^{4} (3 π + α)] - 2 \{\sin^{6} (\frac{π}{2} + α) + \sin^{6} (5 π - α)]$
We know that $\sin (\frac{3 π}{2} - α) = - \cos α, \sin (3 π + α) = - \sin α, \sin (\frac{π}{2} + α) = \cos α$ and $\sin (5 π - α) = \sin α$
Therefore, $3 [\sin^{4} (\frac{3 π}{2} - α) + \sin^{4} (3 π + α)] - 2 \{\sin^{6} (\frac{π}{2} + α) + \sin^{6} (5 π - α)]$
$= 3 [\cos^{4} α + \sin^{4} α] - 2 [\cos^{6} α + \sin^{6} α] = 3 [{(\cos^{2} α)}^{2} + {(\sin^{2} α)}^{2}] - 2 [{(\cos^{2} α)}^{3} + {(\sin^{2} α)}^{3}] = 3 [{(\cos^{2} α + \sin^{2} α)}^{2} - 2 \cos^{2} α \sin^{2} α] - 2 [(\cos^{2} α + \sin^{2} α) (\cos^{4} α + \sin^{4} α - \cos^{2} α \sin^{2} α)] = 3 [1 - 2 \cos^{2} α \sin^{2} α] - 2 [1 - 3 \cos^{2} α \sin^{2} α] = 3 - 6 \cos^{2} α \sin^{2} α - 2 + 6 \cos^{2} α \sin^{2} α = 1$
Hence, the value of the expression is 1.

Page No 54:

Question 23:

If a cos 2θ + b sin 2θ = c has α and β as its roots, then prove that tan α + tan β = $\frac{2 b}{a + c}$ .

Answer:

Given: a cos 2θ + b sin 2θ = c
$\Rightarrow a (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) + b (\frac{2 \tan θ}{1 + \tan^{2} θ}) = c \Rightarrow a - a \tan^{2} θ + 2 b \tan θ = c + c \tan^{2} θ \Rightarrow (c + a) \tan^{2} θ - 2 b \tan θ + (c - a) = 0$
Since, $α$ and $β$ are the roots of the given equation that means $\tan α$ and $\tan β$ can satisfy this equation.
Now, sum of roots = $\frac{- b}{a}$
$\Rightarrow \tan α + \tan β = \frac{- (- 2 b)}{c + a}$
$\Rightarrow \tan α + \tan β = \frac{2 b}{c + a}$ .
Hence proved.

Page No 54:

Question 24:

If x = sec Ï• – tan Ï• and y = cosec Ï• + cot Ï• then show that xy + x – y + 1 = 0

Answer:

Given: $x = s e c ϕ - \tan ϕ$ and $y = \cos e c ϕ + c o t ϕ$
Therefore xy + x – y + 1
$= (s e c ϕ - \tan ϕ) (\cos e c ϕ + c o t ϕ) + (s e c ϕ - \tan ϕ) - (\cos e c ϕ + c o t ϕ) + 1 = (\frac{1 - \sin ϕ}{\cos ϕ}) (\frac{1 + \cos ϕ}{\sin ϕ}) + (\frac{1 - \sin ϕ}{\cos ϕ}) - (\frac{1 + \cos ϕ}{\sin ϕ}) + 1 = \frac{1 + \cos ϕ - \sin ϕ - \sin ϕ \cos ϕ + \sin ϕ - \sin^{2} ϕ - \cos ϕ - \cos^{2} ϕ + \sin ϕ \cos ϕ}{\sin ϕ \cos ϕ} = \frac{0}{\sin ϕ \cos ϕ} = 0$
Hence proved.

Page No 54:

Question 25:

If θ lies in the first quadrant and $\cos θ = \frac{8}{17}$ , then find the value of cos(30° + θ) + cos(45° – θ) + cos(120° – θ).

Answer:

Given: $\cos θ = \frac{8}{17}$ , where θ lies in first quadrant.
Therefore, $\sin θ = \sqrt{1 - \cos^{2} θ} = \sqrt{1 - \frac{64}{289}} = \frac{15}{17}$
Now, $\cos (30 ° + θ) + \cos (45 ° + θ) + \cos (120 ° - θ)$
$= (\cos 30 ° \cos θ - \sin 30 ° \sin θ) + (\cos 45 ° \cos θ + \sin 45 ° \sin θ) + (\cos 120 ° \cos θ + \sin 120 ° \sin θ) = (\frac{\sqrt{3}}{2} \cos θ - \frac{1}{2} \sin θ) + (\frac{1}{\sqrt{2}} \cos θ + \frac{1}{\sqrt{2}} \sin θ) + (- \frac{1}{2} \cos θ + \frac{\sqrt{3}}{2} \sin θ) = (\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} - \frac{1}{2}) \frac{8}{17} + (- \frac{1}{2} + \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2}) \frac{15}{17} = (\frac{1}{17}) \{(\frac{8 \sqrt{3} + 8 \sqrt{2} - 8}{2}) + (\frac{- 15 + 15 \sqrt{2} + 15 \sqrt{3}}{2})\} = \frac{1}{17} (\frac{8 \sqrt{3} + 8 \sqrt{2} - 8 - 15 + 15 \sqrt{2} + 15 \sqrt{3}}{2}) = \frac{23 \sqrt{3} + 23 \sqrt{2} - 23}{34} = \frac{23}{34} (\frac{\sqrt{3} + \sqrt{2} - 1}{2})$

Page No 54:

Question 26:

Find the value of the expression $\cos^{4} \frac{π}{8} + \cos^{4} \frac{3 π}{8} + \cos^{4} \frac{5 π}{8} + \cos^{4} \frac{7 π}{8}$

Answer:

Given: $\cos^{4} \frac{π}{8} + \cos^{4} \frac{3 π}{8} + \cos^{4} \frac{5 π}{8} + \cos^{4} \frac{7 π}{8}$
$= \cos^{4} \frac{π}{8} + \cos^{4} \frac{3 π}{8} + \cos^{4} (π - \frac{3 π}{8}) + \cos^{4} (π - \frac{π}{8}) = \cos^{4} \frac{π}{8} + \cos^{4} \frac{3 π}{8} + \cos^{4} \frac{3 π}{8} + \cos^{4} \frac{π}{8} = 2 (\cos^{4} \frac{π}{8} + \cos^{4} \frac{3 π}{8}) = 2 (\cos^{4} \frac{π}{8} + \sin^{4} (\frac{π}{2} - \frac{3 π}{8})) = 2 (\cos^{4} \frac{π}{8} + \sin^{4} \frac{π}{8})$
$= 2 \{{(\cos^{2} \frac{π}{8} + \sin^{2} \frac{π}{8})}^{2} - 2 \sin^{2} \frac{π}{8} \cos^{2} \frac{π}{8}\} = 2 (1 - 2 \sin^{2} \frac{π}{8} \cos^{2} \frac{π}{8}) = 2 - {(2 \sin \frac{π}{8} \cos \frac{π}{8})}^{2} = 2 - \sin \frac{π}{4} = 2 - \frac{1}{\sqrt{2}} = \frac{2 \sqrt{2} - 1}{\sqrt{2}} = \frac{4 - \sqrt{2}}{2}$

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Question 27:

Find the general solution of the equation 5cos²θ + 7sin²θ – 6 = 0

Answer:

Given: $5 \cos^{2} θ + 7 \sin^{2} θ - 6 = 0$
$\Rightarrow 5 (1 - \sin^{2} θ) + 7 \sin^{2} θ - 6 = 0 \Rightarrow 5 - 5 \sin^{2} θ + 7 \sin^{2} θ - 6 = 0 \Rightarrow 2 \sin^{2} θ - 1 = 0 \Rightarrow \sin θ = \pm \frac{1}{\sqrt{2}}$
When $\sin θ = \frac{1}{\sqrt{2}} = \sin \frac{π}{4}$
$\Rightarrow θ = n π + {(- 1)}^{n} \frac{π}{4}$
When $\sin θ = \frac{- 1}{\sqrt{2}} = \sin (- \frac{π}{4})$
$\Rightarrow θ = n π - {(- 1)}^{n} \frac{π}{4}$
Hence $θ = n π \pm {(- 1)}^{n} \frac{π}{4}, n \in Z$

Page No 55:

Question 28:

Find the general solution of the equation sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x

Answer:

Given: $\sin x - 3 \sin 2 x + \sin 3 x = \cos x - 3 \cos 2 x + \cos 3 x$
$\Rightarrow \sin x + \sin 3 x - 3 \sin 2 x = \cos x + \cos 3 x - 3 \cos 2 x \Rightarrow 2 \sin 2 x \cos x - 3 \sin 2 x = 2 \cos 2 x \cos x - 3 \cos 2 x \Rightarrow \sin 2 x (2 \cos x - 3) = \cos 2 x (2 \cos x - 3) \Rightarrow (2 \cos x - 3) (\sin 2 x - \cos 2 x) = 0$
Either $2 \cos x - 3 = 0 \Rightarrow \cos x = \frac{3}{2}$ , which is not possible.
Or $\sin 2 x = \cos 2 x$
$\Rightarrow \tan 2 x = 1 \Rightarrow \tan 2 x = \tan \frac{π}{4} \Rightarrow 2 x = n π + \frac{π}{4} \Rightarrow x = \frac{n π}{2} + \frac{π}{8}, n \in Z$

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Question 29:

Find the general solution of the equation $(\sqrt{3} - 1) \cos θ + (\sqrt{3} + 1) \sin θ = 2$

Answer:

Given: $(\sqrt{3} - 1) \cos θ + (\sqrt{3} + 1) \sin θ = 2$ .....(1)
Let $r \sin x = \sqrt{3} - 1$ then $r \cos x = \sqrt{3} + 1$
$r^{2} = {(\sqrt{3} - 1)}^{2} + {(\sqrt{3} + 1)}^{2} \Rightarrow r = 2 \sqrt{2}$
And $\frac{r \sin x}{r \cos x} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
$\Rightarrow \tan x = 2 - \sqrt{3} = \tan \frac{π}{12} \Rightarrow x = \frac{π}{12}$
Therefore (1) is,
$r (\sin x \cos θ + \cos x \sin θ) = 2$
$\Rightarrow 2 \sqrt{2} \sin (x + θ) = 2 \Rightarrow \sin (\frac{π}{12} + θ) = \frac{1}{\sqrt{2}} \Rightarrow \sin (\frac{π}{12} + θ) = \sin \frac{π}{4} \Rightarrow θ = n π + {(- 1)}^{n} \frac{π}{4} - \frac{π}{12}$
Hence, $θ = n π + {(- 1)}^{n} \frac{π}{4} - \frac{π}{12}$ , $n \in Z$

Page No 55:

Question 30:

Choose the correct answer from the given four options:
If sin θ + cosec θ = 2, then sin²θ + cosec²θ is equal to
(A) 1
(B) 4
(C) 2
(D) None of these

Answer:

Given: $\sin θ + \cos e c θ = 2$
Squaring both sides,
$\Rightarrow \sin^{2} θ + \cos e c^{2} θ + 2 \sin θ \cos e c θ = 4 \Rightarrow \sin^{2} θ + \cos e c^{2} θ = 4 - 2 = 2$
Hence, the correct answer is option C.

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Question 31:

Choose the correct answer from the given four options:
If f(x) = cos²x + sec²x, then
(A) f(x) < 1
(B) f(x) = 1
(C) 2 < f(x) < 1
(D) f(x) ≥ 2

Answer:

Given: f(x) = cos² x + sec² x
We have AM $\geq$ GM
AM of cos² x and sec² x = $\frac{\cos^{2} x + s e c^{2} x}{2}$
And GM of cos² x and sec² x = $\sqrt{\cos^{2} x s e c^{2} x}$
Therefore, $\frac{\cos^{2} x + s e c^{2} x}{2}$ $\geq$ $\sqrt{\cos^{2} x s e c^{2} x}$
$\Rightarrow \frac{\cos^{2} x + s e c^{2} x}{2}$ $\geq$ 1
$\Rightarrow \frac{f (x)}{2} \geq 1 \Rightarrow f (x) \geq 2$
Hence, the correct answer is option D.

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Question 32:

Choose the correct answer from the given four options:
If $\tan θ = \frac{1}{2} and \tan ϕ = \frac{1}{3}$ , then the value of θ + Ï• is
(A) $\frac{π}{6}$

(B) π

(C) 0

(D) $\frac{π}{4}$

Answer:

Given: $\tan θ = \frac{1}{2} and \tan ϕ = \frac{1}{3}$
Therefore, $\tan (θ + ϕ) = \frac{\tan θ + \tan ϕ}{1 - \tan θ \tan ϕ}$
$= \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1 = \tan \frac{π}{4}$
$\Rightarrow θ + ϕ = \frac{π}{4}$ .
Hence, the correct answer is option D.

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Question 33:

Choose the correct answer from the given four options:
Which of the following is not correct?
(A) $\sin θ = - \frac{1}{5}$

(B) cos θ = 1

(C) $\sec θ = \frac{1}{2}$

(D) tan θ = 20

Answer:

Since the range of sin $θ$ and cos $θ$ is [-1, 1] and the range of tan $θ$ is the set of real numbers R, therefore A, B and D are possible.
The range of Sec $θ$ is $(- \infty, - 1] \cup [1, \infty)$ therefore $s e c θ = \frac{1}{2}$ is not possible.
Hence, the correct answer is option C.

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Question 34:

Choose the correct answer from the given four options:
The value of tan 1° tan 2° tan 3° ... tan 89° is
(A) 0

(B) 1

(C) $\frac{1}{2}$

(D) Not defined

Answer:

Given: $\tan 1 ° \cdot \tan 2 ° \cdot \tan 3 ° \cdot . . . \cdot \tan 89 °$
$= \tan 1 ° \cdot \tan 2 ° \cdot \tan 3 ° \cdot . . . \cdot \tan 45 ° \cdot . . . \cdot \tan 87 ° \cdot \tan 88 ° \cdot \tan 89 ° = (\tan 1 ° \cdot \tan 89 °) \cdot (\tan 2 ° \cdot \tan 88 °) \cdot (\tan 3 ° \cdot \tan 87 °) \cdot . . . \cdot \tan 45 ° = {\tan 1 ° \cdot \tan (90 ° - 1 °)} \cdot {\tan 2 ° \cdot \tan (90 ° - 2 °)} \cdot {\tan 3 ° \cdot \tan (90 ° - 3 °)} \cdot . . . \cdot \tan 45 ° = (\tan 1 ° \cdot c o t 1 °) \cdot (\tan 2 ° \cdot c o t 2 °) \cdot (\tan 3 ° \cdot c o t 3 °) \cdot . . . \cdot \tan 45 ° = 1 \cdot 1 \cdot 1 \cdot . . . . . = 1$
Hence, the correct answer is option B.

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Question 35:

Choose the correct answer from the given four options:
The value of $\frac{1 - \tan^{2} 15 °}{1 + \tan^{2} 15 °}$ is
(A) 1

(B) $\sqrt{3}$

(C) $\frac{\sqrt{3}}{2}$

(D) 2

Answer:

Since $\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ} = \cos 2 θ$
Therefore, $\frac{1 - \tan^{2} 15 °}{1 + \tan^{2} 15 °} = \cos 2 (15 °) = \cos 30 ° = \frac{\sqrt{3}}{2} .$
Hence, the correct answer is option C.

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Question 36:

Choose the correct answer from the given four options:
The value of cos 1° cos 2° cos 3° ... cos 179° is
(A) $\frac{1}{\sqrt{2}}$

(B) 0

(C) 1

(D) –1

Answer:

Given: cos 1° cos 2° cos 3° ... cos 179°
= cos 1° cos 2° cos 3°...cos90°...cos177° cos178° cos179° (cos90° = 0)
= 0.
Hence, the correct answer is option B.

Page No 56:

Question 37:

Choose the correct answer from the given four options:
If tan θ = 3 and θ lies in the third quadrant, then the value of sin θ is
(A) $\frac{1}{\sqrt{10}}$

(B) $- \frac{1}{\sqrt{10}}$

(C) $\frac{- 3}{\sqrt{10}}$

(D) $\frac{3}{\sqrt{10}}$

Answer:

Given: tan θ = 3 and θ lies in the third quadrant.
Therefore, $c o t θ = \frac{1}{3}$
$\Rightarrow \sqrt{\cos e c^{2} θ - 1} = \frac{1}{3} \Rightarrow \cos e c^{2} θ - 1 = \frac{1}{9} \Rightarrow \cos e c^{2} θ = 1 + \frac{1}{9} = \frac{10}{9} \Rightarrow \cos e c θ = \pm \sqrt{\frac{10}{9}} = \pm \frac{\sqrt{10}}{3}$
$\Rightarrow \sin θ = - \frac{3}{\sqrt{10}} (π < θ < \frac{3 π}{2})$ .
Hence, the correct answer is option C.

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Question 38:

Choose the correct answer from the given four options:
The value of tan 75° – cot 75° is equal to
(A) $2 \sqrt{3}$
(B) $2 + \sqrt{3}$
(C) $2 - \sqrt{3}$
(D) 1

Answer:

Given: tan 75° – cot 75°
$= \frac{\sin 75 °}{\cos 75 °} - \frac{\cos 75 °}{\sin 75 °} = \frac{2 (\sin^{2} 75 ° - \cos^{2} 75 °)}{2 \sin 75 ° \cos 75 °} = \frac{- 2 \cos 150 °}{\sin 150 °} = \frac{2 \times \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 2 \sqrt{3}$
Hence, the correct answer is option A.

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Question 39:

Choose the correct answer from the given four options:
Which of the following is correct?
(A) sin1° > sin 1
(B) sin 1° < sin 1
(C) sin 1° = sin 1
(D) $\sin 1 ° = \frac{π}{18 °} \sin 1$

Answer:

Since $π rad = 180 °$
$∴ 1 rad = (\frac{180}{3.14}) ° = 57.3 °$
And sine is an increasing function therefore sin 1° < sin 1 is correct.
Hence, the correct answer is option B.

Page No 56:

Question 40:

Choose the correct answer from the given four options:
If $\tan α = \frac{m}{m + 1}, \tan β = \frac{1}{2 m + 1}$ , then α + β is equal to
(A) $\frac{π}{2}$

(B) $\frac{π}{3}$

(C) $\frac{π}{6}$

(D) $\frac{π}{4}$

Answer:

Given: $\tan α = \frac{m}{m + 1}, \tan β = \frac{1}{2 m + 1}$
Therefore, $\tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β}$
$= \frac{\frac{m}{m + 1} + \frac{1}{2 m + 1}}{1 - \frac{m}{m + 1} \cdot \frac{1}{2 m + 1}} = \frac{m (2 m + 1) + (m + 1)}{(m + 1) (2 m + 1) - m} = \frac{2 m^{2} + 2 m + 1}{2 m^{2} + 2 m + 1} = 1 = \tan \frac{π}{4}$
$\Rightarrow α + β = \frac{π}{4}$
Hence, the correct answer is option D.

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Question 41:

Choose the correct answer from the given four options:
The minimum value of 3 cosx + 4 sinx + 8 is
(A) 5
(B) 9
(C) 7
(D) 3

Answer:

Given expression: 3 cosx + 4 sinx + 8
We know that $- \sqrt{a^{2} + b^{2}} \leq a \sin θ + b \cos θ \leq \sqrt{a^{2} + b^{2}}$
$- \sqrt{3^{2} + 4^{2}} \leq 3 \sin x + 4 \cos x \leq \sqrt{3^{2} + 4^{2}} \Rightarrow$
Thus, minimum value of $3 \sin x + 4 \cos x = - \sqrt{3^{2} + 4^{2}} = - 5$
Therefore, the minimum value of 3 cosx + 4 sinx + 8 = -5 + 8 = 3
Hence, the correct answer is option D.

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Question 42:

Choose the correct answer from the given four options:
The value of tan 3A – tan 2A – tan A is equal to
(A) tan 3A tan 2A tan A
(B) – tan 3A tan 2A tan A
(C) tan A tan 2A – tan 2A tan 3A – tan 3A tan A
(D) None of these

Answer:

$\tan 3 A = \tan (A + 2 A) = \frac{\tan A + \tan 2 A}{1 - \tan A \tan 2 A} \Rightarrow \tan 3 A - \tan A \tan 2 A \tan 3 A = \tan A + \tan 2 A \Rightarrow \tan 3 A - \tan 2 A - \tan A = \tan A \tan 2 A \tan 3 A$
Hence, the correct answer is option A.

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Question 43:

Choose the correct answer from the given four options:
The value of sin (45° + θ) – cos (45° – θ) is
(A) 2 cosθ
(B) 2 sinθ
(C) 1
(D) 0

Answer:

Given expression: $\sin (45 ° + θ) - \cos (45 ° - θ)$
$= \sin 45 ° \cos θ + \cos 45 ° \sin θ - \cos 45 ° \cos θ - \sin 45 ° \sin θ = \frac{1}{\sqrt{2}} \cos θ + \frac{1}{\sqrt{2}} \sin θ - \frac{1}{\sqrt{2}} \cos θ - \frac{1}{\sqrt{2}} \sin θ = \frac{1}{\sqrt{2}} (\cos θ + \sin θ - \cos θ - \sin θ) = 0$
Hence, the correct answer is option D.

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Question 44:

Choose the correct answer from the given four options:
The value of $\cot (\frac{π}{4} + θ) \cot (\frac{π}{4} - θ)$ is
(A) –1
(B) 0
(C) 1
(D) Not defined

Answer:

Given expression: $\cot (\frac{π}{4} + θ) \cot (\frac{π}{4} - θ)$
$= \frac{\cos (\frac{π}{4} + θ) \cos (\frac{π}{4} - θ)}{\sin (\frac{π}{4} + θ) \sin (\frac{π}{4} - θ)} = \frac{\cos^{2} \frac{π}{4} - \sin^{2} y}{\sin^{2} \frac{π}{4} - \sin^{2} y} = \frac{\frac{1}{2} - \sin^{2} y}{\frac{1}{2} - \sin^{2} y} = 1$
Hence, the correct answer is option C.

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Question 45:

Choose the correct answer from the given four options:
cos 2θ cos 2Ï• + sin² (θ – Ï•) – sin² (θ + Ï•) is equal to
(A) sin 2(θ + Ï•)
(B) cos 2(θ + Ï•)
(C) sin 2(θ – Ï•)
(D) cos 2(θ – Ï•)

Answer:

Given expression: cos 2θ cos 2Ï• + sin² (θ – Ï•) – sin² (θ + Ï•)
Since $\sin^{2} x - \sin^{2} y = \sin (x + y) \sin (x - y)$
$∴ \sin^{2} (θ - ϕ) - \sin^{2} (θ + ϕ) = \sin 2 θ \sin (- 2 ϕ) = - \sin 2 θ \sin 2 ϕ$
Now,
$\cos 2 θ \cos 2 ϕ + \sin^{2} (θ - ϕ) - \sin^{2} (θ + ϕ) = \cos 2 θ \cos 2 ϕ - \sin 2 θ \sin 2 ϕ = \cos 2 (θ + ϕ)$
Hence, the correct answer is option B.

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Question 46:

Choose the correct answer from the given four options:
The value of cos 12° + cos 84° + cos 156° + cos 132° is
(A) $\frac{1}{2}$

(B) 1

(C) $- \frac{1}{2}$

(D) $\frac{1}{8}$

Answer:

Given expression: cos 12° + cos 84° + cos 156° + cos 132°
= (cos 12° + cos 132°) + (cos 84° + cos 156°)
= 2cos72° cos60° +2cos120° cos36°
= cos72° - cos36°
= cos (90° - 18° ) - cos36°
= sin18° - cos36°
= $= \frac{\sqrt{5} - 1}{4} - \frac{\sqrt{5} + 1}{4} = - \frac{1}{2}$
Hence, the correct answr is option C.

Page No 57:

Question 47:

Choose the correct answer from the given four options:
If $\tan A = \frac{1}{2}, \tan B = \frac{1}{3}$ , then tan (2A + B) is equal to
(A) 1
(B) 2
(C) 3
(D) 4

Answer:

Given: $\tan A = \frac{1}{2}, \tan B = \frac{1}{3}$
$∴ \tan 2 A = \frac{2 \tan A}{1 - \tan^{2} A} = \frac{2 \times \frac{1}{2}}{1 - \frac{1}{4}} = \frac{4}{3}$
Now,
$\tan (2 A + B) = \frac{\tan 2 A + \tan B}{1 - \tan 2 A \tan B} = \frac{\frac{4}{3} + \frac{1}{3}}{1 - \frac{4}{3} \times \frac{1}{3}} = \frac{5}{3} \times \frac{9}{5} = 3$
Hence, the correct answer is option C.

Page No 57:

Question 48:

Choose the correct answer from the given four options:
The value of $\sin \frac{π}{10} \sin \frac{13 π}{10}$ is
(A) $\frac{1}{2}$

(B) $- \frac{1}{2}$

(C) $- \frac{1}{4}$

(D) 1

Answer:

Given: $\sin \frac{π}{10} \sin \frac{13 π}{10}$
$= \sin \frac{π}{10} \sin (π + \frac{3 π}{10}) = - \sin \frac{π}{10} \sin \frac{3 π}{10} = - \sin 18 ° \sin 54 ° = - \sin 18 ° \cos 36 ° = - \frac{\sqrt{5} - 1}{4} \times \frac{\sqrt{5} + 1}{4} = - \{{(\frac{\sqrt{5}}{4})}^{2} - {(\frac{1}{4})}^{2}\} = - \frac{1}{4}$
Hence, the correct answer is option C.

Page No 57:

Question 49:

Choose the correct answer from the given four options:
The value of sin 50° – sin 70° + sin 10° is equal to
(A) 1

(B) 0

(C) $\frac{1}{2}$

(D) 2

Answer:

sin 50° – sin 70° + sin 10° = $2 \cos (\frac{50 ° + 70 °}{2}) \sin (\frac{50 ° - 70 °}{2}) + \sin 10 ° [\sin A - \sin B = 2 \cos (\frac{A + B}{2}) \sin (\frac{A - B}{2})]$
= $- 2 \cos 60 ° \sin 10 ° + \sin 10 °$
= $- 2 (\frac{1}{2}) \sin 10 ° + \sin 10 °$
= 0

Hence, the correct answer is option (B).

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Question 50:

Choose the correct answer from the given four options:
If sin θ + cos θ = 1, then the value of sin 2θ is equal to
(A) 1

(B) $\frac{1}{2}$

(C) 0

(D) –1

Answer:

sin θ + cos θ = 1
⇒ (sin θ + cos θ)² = (1)² [Squaring both sides]
⇒ sin²θ + cos²θ + 2sinθcosθ = 1 [(a + b)² = a² + b² + 2ab]
⇒ 1 + 2sinθcosθ = 1 (sin²θ + cos²θ = 1)
⇒ 2sinθcosθ = 0
⇒ 2sinθ = 0

Hence, the correct answer is option (C).

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Question 51:

Choose the correct answer from the given four options:
If $α + β = \frac{π}{4}$ , then the value of (1 + tan α) (1 + tan β) is
(A) 1
(B) 2
(C) – 2
(D) Not defined

Answer:

$α + β = \frac{π}{4} \Rightarrow \tan (α + β) = \tan \frac{π}{4} \Rightarrow \frac{\tan α + \tan β}{1 - \tan α \tan β} = 1 \Rightarrow \tan α + \tan β = 1 - \tan α \tan β \Rightarrow \tan α + \tan β + \tan α \tan β = 1 \Rightarrow 1 + \tan α + \tan β + \tan α \tan β = 1 + 1 (Adding 1 on both sides) \Rightarrow 1 (1 + \tan α) + \tan β (1 + \tan α) = 2 \Rightarrow (1 + \tan α) (1 + \tan β) = 2$

Hence, the correct answer is option (B).

Page No 58:

Question 52:

Choose the correct answer from the given four options:
If $\sin θ = \frac{- 4}{5} and θ$ lies in third quadrant then the value of $\cos \frac{θ}{2}$ is
(A) $\frac{1}{5}$

(B) $- \frac{1}{\sqrt{10}}$

(C) $- \frac{1}{\sqrt{5}}$

(D) $\frac{1}{\sqrt{10}}$

Answer:

$\begin{array}{rcl} \cos θ & = & \sqrt{1 - \sin^{2} θ} \\ = & \sqrt{1 - {(\frac{- 4}{5})}^{2}} \\ = & \sqrt{1 - \frac{16}{25}} \\ = & \sqrt{\frac{9}{25}} \\ = & \pm \frac{3}{5} \end{array}$

Now,
$\cos θ = 2 \cos^{2} \frac{θ}{2} - 1 (π < θ < \frac{3 π}{2} \Rightarrow \frac{π}{2} < \frac{θ}{2} < \frac{3 π}{4}) \Rightarrow \frac{- 3}{5} = 2 \cos^{2} \frac{θ}{2} - 1 \Rightarrow 2 \cos^{2} \frac{θ}{2} = 1 - \frac{3}{5} \Rightarrow 2 \cos^{2} \frac{θ}{2} = \frac{2}{5} \Rightarrow \cos^{2} \frac{θ}{2} = \frac{1}{5} \Rightarrow \cos \frac{θ}{2} = \pm \frac{1}{\sqrt{5}} \Rightarrow \cos \frac{θ}{2} = - \frac{1}{\sqrt{5}} (∵ \frac{π}{2} < \frac{θ}{2} < \frac{3 π}{4})$

Hence, the correct answer is option (C).

Page No 58:

Question 53:

Choose the correct answer from the given four options:
Number of solutions of the equation tan x + sec x = 2 cosx lying in the interval [0, 2π] is
(A) 0
(B) 1
(C) 2
(D) 3

Answer:

tan x + sec x = 2 cos x
$\Rightarrow \frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x \Rightarrow 1 + \sin x = = 2 \cos^{2} x \Rightarrow 2 \cos^{2} x - \sin x - 1 = 0 \Rightarrow 2 (1 - \sin^{2} x) - \sin x - 1 = 0 \Rightarrow 2 - 2 \sin^{2} x - \sin x - 1 = 0 \Rightarrow - 2 \sin^{2} x - \sin x + 1 = 0 \Rightarrow 2 \sin^{2} x + \sin x - 1 = 0$

Since, the equation is a quadratic equation, the number of solutions will be 2.

Hence, the correct answer is option (C).

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Question 54:

Choose the correct answer from the given four options:
The value of $\sin \frac{π}{18} + \sin \frac{π}{9} + \sin \frac{2 π}{9} + \sin \frac{5 π}{18}$ is given by

(A) $\sin \frac{7 π}{18} + \sin \frac{4 π}{9}$

(B) 1

(C) $\cos \frac{π}{6} + \cos \frac{3 π}{7}$

(D) $\cos \frac{π}{9} + \sin \frac{π}{9}$

Answer:

$\sin \frac{π}{18} + \sin \frac{π}{9} + \sin \frac{2 π}{9} + \sin \frac{5 π}{18} = (\sin \frac{5 π}{18} + \sin \frac{π}{18}) + (\sin \frac{2 π}{9} + \sin \frac{π}{9}) = 2 \sin (\frac{\frac{5 π}{18} + \frac{π}{18}}{2}) \cos (\frac{\frac{5 π}{18} - \frac{π}{18}}{2}) + 2 \sin (\frac{\sin \frac{2 π}{9} + \frac{π}{9}}{2}) \cos (\frac{\sin \frac{2 π}{9} - \frac{π}{9}}{2}) = 2 \sin \frac{π}{6} \cos \frac{π}{9} + 2 \sin \frac{π}{6} \cos \frac{π}{18} = 2 \times \frac{1}{2} \cos \frac{π}{9} + 2 \times \frac{1}{2} \cos \frac{π}{18} = \cos \frac{π}{9} + \cos \frac{π}{18} = \sin (\frac{π}{2} - \frac{π}{9}) + \sin (\frac{π}{2} - \frac{π}{18}) = \sin \frac{7 π}{18} + \sin \frac{8 π}{18} = \sin \frac{7 π}{18} + \sin \frac{4 π}{9}$

Hence, the correct answer is option (A).

Page No 58:

Question 55:

Choose the correct answer from the given four options:
If A lies in the second quadrant and 3tan A + 4 = 0, then the value of 2cot A – 5cos A + sin A is equal to
(A) $\frac{- 53}{10}$

(B) $\frac{23}{10}$

(C) $\frac{37}{10}$

(D) $\frac{7}{10}$

Answer:

3tan A + 4 = 0
⇒ tan A = $- \frac{4}{3}$ (A lies in second quadrant)
⇒ sin A = $\frac{4}{5}$ , cos A = $\frac{- 3}{5}$ and cot A = $\frac{- 3}{4}$ (Using Pythagoras theorem)
⇒ 2cot A – 5cos A + sin A = $2 (\frac{- 3}{4}) - 5 (\frac{- 3}{5}) + \frac{4}{5}$

= $\frac{- 3}{2} + 3 + \frac{4}{5}$

= $\frac{23}{10}$

Hence, the correct answer is option (B).

Page No 58:

Question 56:

Choose the correct answer from the given four options:
The value of cos² 48° – sin² 12° is
(A) $\frac{\sqrt{5} + 1}{8}$

(B) $\frac{\sqrt{5} - 1}{8}$

(C) $\frac{\sqrt{5} + 1}{5}$

(D) $\frac{\sqrt{5} + 1}{2 \sqrt{2}}$

Answer:

cos² 48° – sin² 12° = cos(48° + 12°)cos(48° − 12°) [cos²A – sin²A = cos(A + B)cos(A – B)]
= cos 60° × cos 36°
= $\frac{1}{2} \times \frac{\sqrt{5} + 1}{4}$
= $\frac{\sqrt{5} + 1}{8}$

Hence, the correct answer is option (A).

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Question 57:

Choose the correct answer from the given four options:
If $\tan α = \frac{1}{7}, \tan β = \frac{1}{3}$ , then cos 2α is equal to
(A) sin 2β
(B) sin 4β
(C) sin 3β
(D) cos 2β

Answer:

$\begin{array}{rcl} \cos 2 α & = & \frac{1 - \tan^{2} α}{1 + \tan^{2} α} \\ = & \frac{1 - {(\frac{1}{7})}^{2}}{1 + {(\frac{1}{7})}^{2}} \\ = & \frac{1 - \frac{1}{49}}{1 + \frac{1}{49}} \\ = & \frac{48}{50} \\ = & \frac{24}{25} \end{array}$

Also,
$\begin{array}{rcl} \tan 2 β & = & \frac{2 \tan β}{1 - \tan^{2} β} \\ = & \frac{2 \times \frac{1}{3}}{1 - \frac{1}{9}} \\ = & \frac{\frac{2}{3}}{\frac{8}{9}} \\ = & \frac{3}{4} \end{array}$
Now,
$\begin{array}{rcl} \sin 4 β & = & \frac{2 \tan 2 β}{1 + \tan^{2} 2 β} \\ = & \frac{2 \times \frac{3}{4}}{1 + {(\frac{3}{4})}^{2}} \\ = & \frac{24}{25} \end{array}$

So, cos 2α = sin 4β.

Hence, the correct answer is option (B).

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Question 58:

Choose the correct answer from the given four options:
If $\tan θ = \frac{a}{b}$ , then b cos 2θ + a sin 2θ is equal to
(A) a
(B) b
(C) $\frac{a}{b}$
(D) None

Answer:

$b \cos 2 θ + a \sin 2 θ = b (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) + a (\frac{2 \tan θ}{1 + \tan^{2} θ}) = b (\frac{1 - \frac{a^{2}}{b^{2}}}{1 + \frac{a^{2}}{b^{2}}}) + a (\frac{\frac{2 a}{b}}{1 + \frac{a^{2}}{b^{2}}}) = b (\frac{b^{2} - a^{2}}{b^{2} + a^{2}}) + (\frac{\frac{2 a^{2}}{b}}{\frac{b^{2} + a^{2}}{b^{2}}}) = \frac{b^{3} - a^{2} b}{b^{2} + a^{2}} + \frac{2 a^{2} b}{b^{2} + a^{2}} = \frac{b^{3} + a^{2} b}{b^{2} + a^{2}} = \frac{b (b^{2} + a^{2})}{b^{2} + a^{2}} = b$

Hence, the correct answer is option (B).

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Question 59:

Choose the correct answer from the given four options:
If for real values of x, $\cos θ = x + \frac{1}{x}$ , then
(A) θ is an acute angle
(B) θ is right angle
(C) θ is an obtuse angle
(D) No value of θ is possible

Answer:

$\cos θ = x + \frac{1}{x} \Rightarrow \cos θ = \frac{x^{2} + 1}{x} \Rightarrow x^{2} + 1 - x \cos θ = 0$

For real values of x, the value of discriminant i.e. b² − 4ac ≥ 0.
$\Rightarrow {(- \cos θ)}^{2} - 4 \times 1 \times 1 \geq 0 \Rightarrow \cos^{2} θ - 4 \geq 0 \Rightarrow \cos^{2} θ \geq 4 \Rightarrow \cos θ \geq \pm 2 (- 1 \leq \cos θ \leq 1)$

So, the value of $θ$ is not possible.

Hence, the correct answer is option (D).

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Question 60:

Fill in the blanks
The value of $\frac{\sin 50 °}{\sin 130 °}$ is _________.

Answer:

$\frac{\sin 50 °}{\sin 130 °} = \frac{\sin 50 °}{\sin (180 ° - 50 °)} = \frac{\sin 50 °}{\sin 50 °} = 1$

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Question 61:

Fill in the blanks
If $k = \sin (\frac{π}{18}) \sin (\frac{5 π}{18}) \sin (\frac{7 π}{18})$ , then the numerical value of k is _______.

Answer:

$\begin{array}{rcl} k & = & \sin (\frac{π}{18}) \sin (\frac{5 π}{18}) \sin (\frac{7 π}{18}) \\ = & \sin 10 ° \sin 50 ° \sin 70 ° \\ = & \sin 10 ° \sin (90 ° - 40 °) \sin (90 ° - 20 °) \\ = & \sin 10 ° \cos 40 ° \cos 20 ° \\ = & \sin 10 ° \times \frac{1}{2} (2 \cos 40 ° \cos 20 °) \\ = & \sin 10 ° \times \frac{1}{2} [\cos (40 ° + 20 °) + \cos (40 ° - 20 °)] \\ = & \frac{1}{2} \sin 10 ° (\cos 60 ° + \cos 20 °) \\ = & \frac{1}{2} \sin 10 ° (\frac{1}{2} + \cos 20 °) \\ = & \frac{1}{4} \sin 10 ° + \frac{1}{2} \sin 10 ° \cos 20 ° \\ = & \frac{1}{4} \sin 10 ° + \frac{1}{4} (2 \sin 10 ° \cos 20 °) \\ = & \frac{1}{4} \sin 10 ° + \frac{1}{4} [\sin (10 ° + 20 °) + \sin (10 ° - 20 °)] \\ = & \frac{1}{4} \sin 10 ° + \frac{1}{4} (\sin 30 ° + \sin (- 10 °)) \\ = & \frac{1}{4} \sin 10 ° + \frac{1}{4} \sin 30 ° - \frac{1}{4} \sin 10 ° \\ = & \frac{1}{4} \times \frac{1}{2} \\ = & \frac{1}{8} \end{array}$

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Question 62:

Fill in the blanks
If $\tan A = \frac{1 - \cos B}{\sin B}$ , then tan 2A = __________.

Answer:

$\begin{array}{rcl} \tan 2 A & = & \frac{2 \tan A}{1 - \tan^{2} A} \\ = & \frac{2 (\frac{1 - \cos B}{\sin B})}{1 - {(\frac{1 - \cos B}{\sin B})}^{2}} \\ = & \frac{2 (\frac{2 \sin^{2} \frac{B}{2}}{2 \sin \frac{B}{2} \cos \frac{B}{2}})}{1 - {(\frac{2 \sin^{2} \frac{B}{2}}{2 \sin \frac{B}{2} \cos \frac{B}{2}})}^{2}} \\ = & \frac{2 \frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}}{1 - {(\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}})}^{2}} \\ = & \frac{2 \tan \frac{B}{2}}{1 - \tan^{2} \frac{B}{2}} \\ = & \tan B \end{array}$

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Question 63:

Fill in the blanks
If sin x + cos x = a, then
(i) sin⁶x + cos⁶x = _______
(ii) |sin x – cos x| = _______.

Answer:

$\sin x + \cos x = a \Rightarrow {(\sin x + \cos x)}^{2} = a^{2} \Rightarrow \sin^{2} x + \cos^{2} x + 2 \sin x \cos x = a^{2} \Rightarrow 1 + 2 \sin x \cos x = a^{2} \Rightarrow \sin x \cos x = \frac{a^{2} - 1}{2}$

(i)
$\sin^{6} x + \cos^{6} x = {(\sin^{2} x)}^{3} + {(\cos^{2} x)}^{3} = {(\sin^{2} x + \cos^{2} x)}^{3} - 3 \sin^{2} x \cos^{2} x (\sin^{2} x + \cos^{2} x) = 1^{3} - 3 {(\frac{a^{2} - 1}{2})}^{2} \times 1 = 1 - \frac{3 {(a^{2} - 1)}^{2}}{4} = \frac{1}{4} [4 - 3 {(a^{2} - 1)}^{2}]$

(ii)
${|\sin x - \cos x|}^{2} = \sin^{2} x + \cos^{2} x - 2 \sin x \cos x = 1 - 2 (\frac{a^{2} - 1}{2}) = 1 - a^{2} + 1 = 2 - a^{2}$
$\Rightarrow |\sin x - \cos x| = \sqrt{2 - a^{2}}$ (Since, |sin x – cos x| > 0)

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Question 64:

Fill in the blanks
In a triangle ABC with ∠C = 90° the equation whose roots are tan A and tan B is ________.

Answer:

$A + B = 90 ° (∠ C = 90 °) \Rightarrow \tan (A + B) = \tan 90 ° \Rightarrow \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{1}{0} \Rightarrow 1 - \tan A \tan B = 0 \Rightarrow \tan A \tan B = 1$

Also,
$\begin{array}{rcl} \tan A + \tan B & = & \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} \\ = & \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B} \\ = & \frac{\sin (A + B)}{\cos A \cos B} \\ = & \frac{\sin 90 °}{\cos A \cos (90 ° - A)} \\ = & \frac{1}{\cos A \sin A} \\ = & \frac{2}{\sin 2 A} \end{array}$

The roots of the equation tan A and tan B. So,
$x^{2} - (\tan A + \tan B) x + \tan A \tan B = 0 \Rightarrow x^{2} - (\frac{2}{\sin 2 A}) x + 1 = 0$

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Question 65:

Fill in the blanks
3(sin x – cos x)⁴ + 6(sin x + cos x)² + 4(sin⁶x + cos⁶x) = _______.

Answer:

$3 {(\sin x - \cos x)}^{4} + 6 {(\sin x + \cos x)}^{2} + 4 (\sin 6 x + \cos 6 x) = 3 {(\sin^{2} x + \cos^{2} x - 2 \sin x \cos x)}^{2} + 6 (\sin^{2} x + \cos^{2} x + 2 \sin x \cos x) + 4 [{(\sin^{2} x)}^{3} + {(\cos^{2} x)}^{3}] = 3 {(1 - 2 \sin x \cos x)}^{2} + 6 (1 + 2 \sin x \cos x) + 4 [{(\sin^{2} x + \cos^{2} x)}^{3} - 3 \sin^{2} x \cos^{2} x (\sin^{2} x + \cos^{2} x)] = 3 (1 + 4 \sin^{2} x \cos^{2} x - 4 \sin x \cos x) + 6 + 12 \sin x \cos x + 4 - 12 \sin^{2} x \cos^{2} x = 3 + 12 \sin^{2} x \cos^{2} x - 12 \sin x \cos x + 6 + 12 \sin x \cos x + 4 - 12 \sin^{2} x \cos^{2} x = 3 + 6 + 4 = 13$

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Question 66:

Fill in the blanks
Given x > 0, the values of f(x) = – 3 cos $\sqrt{3 + x + x^{2}}$ lie in the interval _______.

Answer:

Let $\sqrt{3 + x + x^{2}}$ = y.
Then,
f(x) = –3cos y
Now,
$- 1 \leq \cos y \leq 1 \Rightarrow 3 \geq - 3 \cos y \geq - 3 \Rightarrow - 3 \leq - 3 \cos y \leq 3 \Rightarrow - 3 \leq - 3 \cos \sqrt{3 + x + x^{2}} \leq 3, x > 0$

Hence, given x > 0, the values of f(x) = – 3 cos $\sqrt{3 + x + x^{2}}$ lie in the interval [−3, 3].

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Question 67:

Fill in the blanks
The maximum distance of a point on the graph of the function $y = \sqrt{3}$ sin x + cos x from x-axis is _______.

Answer:

We have,
$y = \sqrt{3}$ sin x + cos x
$= 2 [\frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x] = 2 [\cos \frac{π}{6} \sin x + \sin \frac{π}{6} \cos x] = 2 \sin (x + \frac{π}{6})$
Thus, Maximum value of $y = \sqrt{3}$ sin x + cos x is 2.

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Question 68:

State whether the following statement is True or False.
If $\tan A = \frac{1 - \cos B}{\sin B}$ , then tan 2A = tan B

Answer:

Given that, $\tan A = \frac{1 - \cos β}{\sin β}$
Now, taking LHS
$\tan 2 A = \frac{2 \tan A}{1 - \tan^{2} β}$

$= [\frac{[2 \frac{(1 - \cos β)}{\sin β}]}{1 - {(\frac{1 - \cos β}{\sin β})}^{2}}] = [\frac{\frac{\frac{2 (1 - \cos β)}{\sin β}}{\sin^{2} β - (1 + \cos^{2} β - 2 \cos β)}}{\sin^{2} β}] = \frac{2 (1 - \cos β) \sin β}{\sin^{2} β - (1 + \cos^{2} β - 2 \cos β)} = \frac{2 (1 - \cos β) \sin β}{[\sin^{2} β - 1 + \cos^{2} β + 2 \cos β]} = \frac{2 (1 - \cos β) \sin β}{[1 - \cos^{2} β - 1 - \cos^{2} β + 2 \cos β]} = \frac{2 (1 - \cos β) \sin β}{2 \cos β (1 - \cos β)} = \tan β = RHS$

Thus, True.

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Question 69:

State whether the following statement is True or False.
The equality sin A + sin 2A + sin 3A = 3 holds for some real value of A.

Answer:

Given: sin A + sin 2A + sin 3A = 3.
Since, the maximum value of sin A is 1 but for sin2 A and sin 3A it is not equal to 1.
Thus, the given equation is False.

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Question 70:

State whether the following statement is True or False.
sin 10° is greater than cos 10°.

Answer:

Let sin 10° > cos 10°
⇒ sin 10° > cos(90° – 80°)
⇒ sin 10° > sin 80°
Which is wrong as for sin, value increase with θ.
Hence, False.

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Question 71:

State whether the following statement is True or False.
$\cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{8 π}{15} \cos \frac{16 π}{15} = \frac{1}{16}$

Answer:

Taking LHS, we have
$\Rightarrow \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{8 π}{15} \cos \frac{16 π}{15} \Rightarrow \frac{(2 \sin \frac{2 π}{15} \cos \frac{2 π}{15}) \cdot \cos \frac{4 π}{15} \cdot \cos \frac{8 π}{15} \cdot \cos \frac{16 π}{15}}{2 \cdot \cos \frac{2 π}{15}} \Rightarrow \frac{\sin \frac{4 π}{15} \cos \frac{4 π}{15} \cdot \cos \frac{8 π}{15} \cos \frac{16 π}{15}}{2 \cos \frac{2 π}{15}} \Rightarrow \frac{(2 \sin \frac{4 π}{15} \cos \frac{4 π}{15}) \cos \frac{8 π}{15} \cdot \cos \frac{16 π}{15}}{4 \sin \frac{2 π}{15} .} \Rightarrow \frac{\sin \frac{8 π}{15} \cos \frac{8 π}{15} \cos \frac{16 π}{15}}{4 \sin \frac{2 π}{15}} \Rightarrow \frac{(2 \sin \frac{8 π}{15} \cos \frac{8 π}{15}) \cos \frac{16 π}{16}}{8 \sin \frac{2 π}{15}} \Rightarrow \frac{\sin \frac{16 π}{15} \cos \frac{16 π}{15}}{8 \sin \frac{2 π}{15}} \Rightarrow \frac{2 \sin \frac{16 π}{15} \cos \frac{16 π}{15}}{16 \sin \frac{2 π}{15}} \Rightarrow \frac{\sin (2 π + \frac{2 π}{15})}{16 \sin \frac{2 π}{15}} = \frac{1}{16} = RHS$

Hence, Ture

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Question 72:

State whether the following statement is True or False.
One value of θ which satisfies the equation sin⁴θ – 2sin²θ – 1 lies between 0 and 2π.

Answer:

Given that, sin⁴ θ – 2sin² θ – 1
Now,
$\begin{array}{rcl} \sin^{2} θ & = & \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 \times 1 \times (- 1)}}{2 \times 1} \\ = & \frac{2 \pm 2 \sqrt{2}}{2} \\ = & 1 \pm \sqrt{2} \end{array}$
$\Rightarrow \sin^{2} θ = 1 + \sqrt{2} or 1 - \sqrt{2} . But, \sin^{2} θ \leq 1 Since \sin^{2} θ = 1 + \sqrt{2} or 1 - \sqrt{2} .$
Which is not possible.

Hence, about statement is False.

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Question 73:

State whether the following statement is True or False.
If cosec x = 1 + cot x then x = 2nπ, 2nπ + $\frac{π}{2}$

Answer:

Given that,
cosec x = 1 + cot x
$\Rightarrow \frac{1}{\sin x} = 1 + \frac{\cos x}{\sin x} \Rightarrow \frac{1}{\sin x} = \frac{\sin x \cos x}{\sin x} \Rightarrow \sin x + \cos x = 1 \Rightarrow \frac{1}{\sqrt{2}} \sin + \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}} \Rightarrow \sin \frac{π}{4} \sin x + \cos \frac{π}{4} \cos x = \frac{1}{\sqrt{2}} \Rightarrow \cos (x - \frac{π}{4}) = \cos \frac{π}{4} \Rightarrow x = 2 n π + \frac{π}{4} + \frac{π}{4} = 2 n π + \frac{π}{2} OR \Rightarrow x = 2 n π + \frac{π}{4} - \frac{π}{4} \Rightarrow 2 n π$

Hence, True.

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Question 74:

State whether the following statement is True or False.
If $\tan θ + \tan 2 θ + \sqrt{3} \tan θ \tan 2 θ = \sqrt{3}, then θ = \frac{n π}{3} + \frac{π}{9}$

Answer:

Given that,
$\tan θ + \tan 2 θ + \sqrt{3} \tan θ \tan 2 θ = \sqrt{3} .$
$\Rightarrow \tan θ + \tan 2 θ + \sqrt{3} (1 - \tan θ \tan 2 θ) . \Rightarrow \frac{\tan θ + \tan 2 θ}{1 - \tan θ \tan 2 θ} = \sqrt{3} . \Rightarrow \tan 3 θ + \sqrt{3} = \tan \frac{π}{3} . \Rightarrow 3 θ = n π + \frac{π}{3} n ε Z . θ = \frac{n π}{3} + \frac{π}{9}$

Hence, True.

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Question 75:

State whether the following statement is True or False.
If tan (π cos θ) = cot (π sin θ), then $\cos (θ - \frac{π}{4}) = \pm \frac{1}{2 \sqrt{2}}$

Answer:

Given that,
tan (π cos θ) = cot (π sin θ),
$\Rightarrow πcosx = \frac{π}{2} - πsinx . \Rightarrow cosx + sinx = \frac{π}{2} \Rightarrow \frac{1}{\sqrt{2}} cosx + \frac{1}{\sqrt{2}} sinx = \frac{1}{2 \sqrt{2}} . \Rightarrow \cos (x - \frac{π}{4}) = \frac{1}{2 \sqrt{2}}$

Hence, True.

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Question 76:

State whether the following statement is True or False.

In the following match each item given under the column C₁ to its correct answer given under the column C₂ :

(a) sin (x + y) sin (x – y)	(i) cos²x – sin²y
(b) cos (x + y) cos (x – y)	(ii) $\frac{1 - \tan θ}{1 + \tan θ}$
(c) $\cot (\frac{π}{4} + θ)$	(iii) $\frac{1 + \tan θ}{1 - \tan θ}$
(d) $\tan (\frac{π}{4} + θ)$	(iv) sin²x – sin²y

Answer:

We have,
sin (x + y) sin (x – y) = sin² x – sin² y
cos (x + y) cos (x – y) = cos² x – sin² y
$\begin{array}{rcl} \cot (\frac{π}{4} + θ) & = & \frac{\cot \frac{π}{4} \cot θ - 1}{\cot θ + \cot \frac{π}{4}} \\ = & \frac{\cot θ - 1}{\cot θ + 1} \\ = & \frac{1 - \tan θ}{1 + \tan θ} \end{array}$

$\begin{array}{rcl} \tan (\frac{π}{4} + θ) & = & \frac{\tan \frac{π}{4} \tan θ + 1}{\tan θ + \tan} \\ = & \frac{1 + \tan θ}{1 - \tan θ} \end{array}$
Thus, a-(iv), b-(i), c-(ii), d-(iii)

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