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#### Question 1:

sin 60° cos 30° + cos 60° sin 30°

On substituting the values of various T-ratios, we get:
sin 60o cos 30o + cos 60o sin 30o
$=\left(\frac{\sqrt{3}}{2}×\frac{\sqrt{3}}{2}+\frac{1}{2}×\frac{1}{2}\right)=\left(\frac{3}{4}+\frac{1}{4}\right)=\frac{4}{4}=1$

#### Question 2:

cos 60° cos 30° − sin 60° sin 30°

On substituting the values of various T-ratios, we get:
cos 60o cos 30o − sin 60o sin 30o
$=\left(\frac{1}{2}×\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}×\frac{1}{2}\right)=\left(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\right)=0$

#### Question 3:

cos 45° cos 30° + sin 45° sin 30°

On substituting the values of various T-ratios, we get:
cos 45o cos 30o + sin 45o  sin 30o

#### Question 4:

2 cos2 60° + 3 sin2 45° − 3 sin2 30° + 2 cos2 90°

On substituting the values of various T-ratios, we get:
2 cos2 60o + 3 sin2 45o − 3 sin2 30o + 2 cos2 90o

#### Question 5:

cot230° − 2cos230° − $\frac{3}{4}$ sec245° + $\frac{1}{4}$ cosec230°

On substituting the values of various T-ratios, we get:
cot2 30o − 2 cos2 30o − $\frac{3}{4}$ sec2 45o + $\frac{1}{4}$ cosec2 30o

#### Question 6:

(sin230° + 4cot245° − sec260°)(cosec245° sec230°)

On substituting the values of various T-ratios, we get:
(sin2 30o + 4 cot2 45o − sec2 60o )(cosec2 45o sec2 30o)

#### Question 7:

$\frac{4}{{\mathrm{cot}}^{2}30°}+\frac{1}{{\mathrm{sin}}^{2}30°}-2{\mathrm{cos}}^{2}45°-{\mathrm{sin}}^{2}0°$

On substituting the values of various T-ratios, we get:

#### Question 8:

$\frac{{\mathrm{tan}}^{2}60°+4{\mathrm{cos}}^{2}45°+3{\mathrm{cosec}}^{2}60°+2{\mathrm{cos}}^{2}90°}{2\mathrm{cosec}30°+3\mathrm{sec}60°-\frac{7}{3}{\mathrm{cot}}^{2}30°}$

On substituting the values of various T-ratios, we get:

#### Question 9:

$\frac{\mathrm{sin}30°}{\mathrm{cos}45°}+\frac{\mathrm{cot}45°}{\mathrm{sec}60°}-\frac{\mathrm{sin}60°}{\mathrm{tan}45°}-\frac{\mathrm{cos}30°}{\mathrm{sin}90°}$

On substituting the values of various T-ratios, we get:

#### Question 10:

Show that:
(i) $\frac{1-\mathrm{sin}60°}{\mathrm{cos}60°}=\frac{\mathrm{tan}60°-1}{\mathrm{tan}60°+1}$
(ii) $\frac{\mathrm{cos}30°+\mathrm{sin}60°}{1+\mathrm{sin}30°+\mathrm{cos}60°}=\mathrm{cos}30°$

(i)

Hence, LHS = RHS

(ii)

Hence, LHS = RHS

1sin60°cos60°

#### Question 11:

Verify each of the following:
(i) sin 60° cos 30° − cos 60° sin 30° = sin 30°
(ii) cos 60° cos 30° + sin 60° sin 30° = cos 30°
(iii) 2 sin 30° cos 30° = sin 60°
(iv) 2 sin 45° cos 45° = sin 90°

(i) sin 60o cos 30o − cos 60o sin 30o

∴ sin 60o cos 30o − cos 60o sin 30o = sin 30o

(ii) cos 60o cos 30o + sin 60o sin 30o

∴ cos 60o cos 30o + sin 60o sin 30o = cos 30o

(iii) 2 sin 30o cos 30o

∴ 2 sin 30o cos 30o = sin 60o

(iv) 2 sin 45o cos 45o
$=2×\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}=1$
Also, sin 90o = 1
∴ 2 sin 45o cos 45o = sin 90o

#### Question 12:

If A = 45°, verify that:
(i) sin 2A = 2 sin A cos A
(ii) cos 2A = 2 cos2 A − 1 = 1 − 2 sin2 A

A = 45o
⇒ 2A = 2 $×$ 45o = 90o

(i) sin 2A = sin 90o = 1
2 sin A cos A = 2 sin 45o cos 45o
∴ sin 2A = 2 sin A cos A

(ii) cos 2A = cos 90o = 0
2 cos2 A − 1 = 2 cos2 45o − 1 =
Now, 1 − 2 sin2 A
∴ cos 2A 2 cos2 A − 1 = 1 − 2 sin2 A

#### Question 13:

If A = 30°, verify that:
(i) $\mathrm{sin}2A=\frac{2\mathrm{tan}A}{1+{\mathrm{tan}}^{2}A}$
(ii) $\mathrm{cos}2A=\frac{1-{\mathrm{tan}}^{2}A}{1+{\mathrm{tan}}^{2}A}$
(iii) $\mathrm{tan}2A=\frac{2\mathrm{tan}A}{1-{\mathrm{tan}}^{2}A}$

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

​(i) sin 2A = sin 60o = $\frac{\sqrt{3}}{2}$

(ii) cos 2A = cos 60o = $\frac{1}{2}$

(iii) tan 2A = tan 60o = $\sqrt{3}$

=2tanA1+tan2A

#### Question 14:

Using the formula,  , find the value of cos 30°, it being given that cos 60° = $\frac{1}{2}$.

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ cos 30o = $\frac{\sqrt{3}}{2}$

#### Question 15:

Using the formula,  , find the value of sin 30°, it being given that cos 60° = $\frac{1}{2}$.

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ sin 30o = $\frac{1}{2}$

#### Question 16:

Using the formula, tan , find the value of tan 60°, it being given that tan 30° = $\frac{1}{\sqrt{3}}$.

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ tan 60o = $\sqrt{3}$

#### Question 18:

If A = 60° and B = 30°, verify that:
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) cos (A + B) = cos A cos B − sin A sin B

A = 60o and B = 30o
Now, A + B = 60o + 30o​ = 90o
Also, A − B = 60o − 30o = 30o

(i) sin (A + B) = sin 90o = 1
sin A cos B + cos A sin B = sin 60o cos 30o + cos 60o sin 30o

∴ sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos 90o = 0
cos A cos B − sin A sin B =cos 60o cos 30o − sin 60o sin 30o
∴​ cos (A + B) = cos A cos B − sin A sin B

#### Question 19:

If A and B are acute angles such that tan A = $\frac{1}{3}$, tan B$\frac{1}{2}$ and tan (A + B) =  , show that A + B = 45°.

Given:

13

#### Question 20:

In the adjoining figure, ∆ABC is a right-angled triangle in which ∠B = 90°, ∠A = 30° and AC = 20 cm.
Find (i) BC, (ii) AB.
Figure

From the given right-angled triangle, we have:

#### Question 17:

If A = B = 45°, verify that:
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) sin (A B) = sin A cos B − cos A sin B
(iii) cos (A + B) = cos A cos B − sin A sin B
(iv) cos (AB) = cos A cos B + sin A sin B

A = B = 45o
Now, A + B = 45o + 45o​ = 90o
Also, AB = 45o − 45o = 0o

(i) sin (A + B) = sin 90o = 1
sin A cos B + cos A sin B = sin 45o cos 45o + cos 45o sin 45o
=
∴ sin (A + B) = sin A cos B + cos A sin B

(ii) sin (AB) = sin 0o = 0
sin A cos B − cos A sin B = sin 45o cos 45o − cos 45o sin 45o
=
∴ sin (AB) = sin A cos B − cos A sin B

(iii) cos (A + B) = cos 90o = 0
cos A cos B − sin A sin B = cos 45o cos 45o − sin 45o sin 45o
=
∴​ cos (A + B) = cos A cos B − sin A sin B

(iv) cos (AB) = cos 0o = 1
cos A cos B + sin A sin B = cos 45o cos 45o + sin 45o sin 45o
=
∴​ cos (AB) = cos A cos B + sin A sin B

#### Question 21:

In the adjoining figure, ∆ ABC is a right-angled at B and ∠A = 30°. If BC = 6 cm,
Find (i) AB, (ii) AC.
Figure

From the given right-angled triangle, we have:

#### Question 22:

In the adjoining figure, ∆ABC is a right-angled at B and ∠A = 45°. If AC$3\sqrt{2}$ cm,
find (i) BC, (ii) AB.
Figure

From right-angled ∆ABC, we have:

#### Question 23:

If sin (A + B) = 1 and cos (A − B) = 1, 0° ≤ (A + B) ≤ 90° and A > B, then find A and B.

Here, sin (A + B) = 1
⇒ sin (A+ B) = sin 90o                     [∵ sin 90o = 1]
A + B = 90o​                       ...(i)

Also, cos (AB) = 1
⇒​ cos (A − B) = cos 0o                    [∵​ cos 0o = 1]
A B = 0o                        ...(ii)

Solving (i) and (ii), we get:
A = 45o and B = 45o

#### Question 24:

If sin (AB) = $\frac{1}{2}$ and cos (A + B) = $\frac{1}{2}$, 0° < (A + B) < 90° and A > B, then find A and B.

Here, sin (A − B) = $\frac{1}{2}$
⇒ sin (A B) = sin 30o                [∵ sin 30o = $\frac{1}{2}$]
A − B = 30o​                                  ...(i)

Also, cos (A + B) = $\frac{1}{2}$
⇒​ cos (A + B) =  cos 60o              [∵​ cos 60o = $\frac{1}{2}$]
A + B = 60o                                 ...(ii)

Solving (i) and (ii), we get:
A = 45o and B = 15o

#### Question 25:

If tan (A − B) = $\frac{1}{\sqrt{3}}$ and tan (A + B) = $\sqrt{3}$, 0° < (A + B) < 90° and A > B, then find A and B.

Here, tan (A − B) = $\frac{1}{\sqrt{3}}$
⇒ tan (A B) = tan 30o       [∵ tan 30o = $\frac{1}{\sqrt{3}}$]
A − B = 30o                     ...(i)

Also, tan (A + B) = $\sqrt{3}$
⇒​ tan (A + B) =  tan 60o        [∵ tan 60o = $\sqrt{3}$]
A + B = 60o                           ...(ii)

Solving (i) and (ii), we get:
A = 45o and B = 15o

#### Question 26:

Find the value of sin 30° geometrically.

Consider an equilateral ∆ABC with each side equal to 2a.
Then, each angle of ​∆ABC = 60o
Figure
Then, it is clear that BD = DC = a
⇒ 90o + 60o + ∠BAD = 180o
⇒ ∠BAD = 180o − 150o
∴ sin 30o

#### Question 27:

Find the value of sin 60° geometrically.

Consider an equilateral ∆ABC with each side equal to 2a.
Then, each angle of ​∆ABC = 60o
Figure
Then, it is clear that BD = DC = a
Now, in  ​∆ ADB, we have:
⇒ 90o + 60o + ∠ BAD = 180o
⇒ ∠BAD = 180o − 150o = 30o

In ∆ADB, by Pythagoras theorem, we have:

∴ sin 60o

#### Question 1:

(sec260°− 1) = ?
(a) 2
(b) 3
(c) 4
(d) 0

(b) 3

sec260o − 1 = (2)2 − 1 = 4 − 1 = 3

#### Question 2:

If x tan 45° cos 60° = sin 60° cot 60°, then x = ?
(a) 1
(b) $\frac{1}{2}$
(c) $\frac{1}{\sqrt{2}}$
(d) $\sqrt{3}$

(a) 1

#### Question 3:

If (tan245° − cos230°) = x sin 45° cos 45°, then x = ?
(a) 2
(b) −2
(c) $\frac{1}{2}$
(d) $\frac{-1}{2}$

(c) $\frac{1}{2}$

#### Question 4:

If tan x = 3 cot x, then x = ?
(a) 45°
(b) 60°
(c) 30°
(d) 15°

(b) 60o

#### Question 5:

If 2 sin 2θ = $\sqrt{3}$, then θ = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

(a) 30o

#### Question 6:

If 2cos 3θ = 1, then θ = ?
(a) 10°
(b) 15°
(c) 20°
(d) 30°

(c) 20o

#### Question 7:

If $\sqrt{3}$ tan 2θ − 3 = 0, then θ = ?
(a) 15°
(b) 30°
(c) 45°
(d) 60°

(b) 30o

#### Question 8:

If sin (A + B) = 1 and cos (A − B) = $\frac{\sqrt{3}}{2}$, where A > B, then B = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

(a) 30o

#### Question 9:

If tan (A + B) = $\sqrt{3}$ and tan (AB) = $\frac{1}{\sqrt{3}}$, where A ≥ B and (A + B) is acute, then A = ?
(a) 15°
(b) 30°
(c) 45°
(d) 60°

(c) 45o

#### Question 10:

In ∆ ABC, ∠B = 90°, AB = 3 cm and AC = 6 cm.
Then, ∠A = ?
(a) 30°
(b) 60°
(c) 45°
(d) None of these
Figure

(b) 60o

In right-angled ∆ ABC, we have:

#### Question 11:

In ∆ABC, ∠B = 90°, ∠A = 30° and AB = 9 cm.
Then, BC = ?
(a) 3 cm
(b)
(c)
(d) 6 cm
Figure

(c )

In right-angled ∆ ABC, we have:

#### Question 12:

The value of (cos 60° cos 30° − sin 60° sin 30°) is
(a) 0
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{1}{2}$
(d) 1

(a) 0

#### Question 13:

(sin 60° cos 30° − cos 60° sin 30°) = ?
(a) 0
(b) 1
(c) $\frac{1}{2}$
(d) $\frac{\sqrt{3}}{2}$

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

#### Question 14:

(tan 30° cosec 60° + tan 60° sec 30°) = ?
(a) $2\frac{1}{3}$
(b) $2\frac{2}{3}$
(c) $3\frac{1}{3}$
(d) $1\frac{2}{3}$

(b ) $2\frac{2}{3}$

#### Question 15:

(cos 0° + sin 30° + sin 45°) (sin 90° + cos 60° − cos 45°) = ?
(a) $\frac{5}{6}$
(b) $\frac{5}{8}$
(c) $\frac{3}{5}$
(d) $\frac{7}{4}$

(d) $\frac{7}{4}$

#### Question 16:

(sin230° + 4cot245° − sec260°) = ?
(a) 0
(b) $\frac{1}{4}$
(c) 4
(d) 1

(b) $\frac{1}{4}$

#### Question 17:

(3cos260° + 2cot230° − 5sin245°) = ?
(a) $\frac{13}{6}$
(b) $\frac{17}{4}$
(c) 1
(d) 4

(b) $\frac{17}{4}$

#### Question 18:

${\mathrm{cos}}^{2}30°{\mathrm{cos}}^{2}45°+4{\mathrm{sec}}^{2}60°+\frac{1}{2}{\mathrm{cos}}^{2}90°-2{\mathrm{tan}}^{2}60°=?$
(a) $\frac{73}{8}$
(b) $\frac{75}{8}$
(c) $\frac{81}{8}$
(d) $\frac{83}{8}$

(d) $\frac{83}{8}$

#### Question 19:

$\left({\mathrm{sin}}^{2}30°{\mathrm{cos}}^{2}45°+4{\mathrm{tan}}^{2}30°+\frac{1}{2}{\mathrm{sin}}^{2}90°+\frac{1}{8}{\mathrm{cot}}^{2}60°\right)=?$
(a) 2
(b) 3
(c) 4
(d) $\frac{5}{2}$

(a) 2

#### Question 20:

$\left(\frac{4}{3}{\mathrm{cot}}^{2}30°+3{\mathrm{sin}}^{2}60°-2\mathrm{cos}e{c}^{2}60°-\frac{3}{4}{\mathrm{tan}}^{2}30°\right)=?$
(a) 3
(b) $\frac{8}{3}$
(c) $\frac{10}{3}$
(d) $\frac{9}{4}$

(c) $\frac{10}{3}$

#### Question 21:

$\frac{\left({\mathrm{tan}}^{2}60°+4{\mathrm{sin}}^{2}45°+3{\mathrm{sec}}^{2}30°+5{\mathrm{cos}}^{2}90°\right)}{\left(\mathrm{cosec}30°+\mathrm{sec}60°-{\mathrm{cot}}^{2}30°\right)}=?$
(a) $\frac{9}{7}$
(b) 9
(c) $\frac{7}{9}$
(d) 6

(b) 9

#### Question 22:

$\frac{1}{4}\left({\mathrm{cot}}^{4}30°-{\mathrm{cosec}}^{4}60°\right)+\frac{3}{2}\left({\mathrm{sec}}^{2}45°-{\mathrm{tan}}^{2}30°\right)-5{\mathrm{cos}}^{2}60°=?$
(a) $\frac{35}{9}$

(b) $\frac{45}{8}$

(c) $\frac{55}{18}$

(d) $\frac{49}{36}$

(c) $\frac{55}{18}$

#### Question 23:

$4\left({\mathrm{sin}}^{4}30°+{\mathrm{cos}}^{4}60°\right)-\frac{2}{3}\left({\mathrm{sin}}^{2}60°-{\mathrm{cos}}^{2}45°\right)+\frac{1}{2}{\mathrm{tan}}^{2}60°=?$
(a) $\frac{8}{3}$
(b) $\frac{11}{6}$
(c) $\frac{13}{6}$
(d) $\frac{13}{8}$

(b) $\frac{11}{6}$

#### Question 24:

$\frac{2}{3}\left({\mathrm{cos}}^{4}30°-{\mathrm{sin}}^{4}45°\right)-3\left({\mathrm{sin}}^{2}60°-{\mathrm{sec}}^{2}45°\right)+\frac{1}{4}{\mathrm{cot}}^{2}30°=?$
(a) $\frac{53}{12}$
(b) $\frac{73}{24}$
(c) $\frac{113}{24}$
(d) $\frac{83}{12}$

(c) $\frac{113}{24}$

#### Question 25:

If sin α = $\frac{1}{2}$ and cos β = $\frac{1}{2}$, then (α + β) = ?
(a) 0°
(b) 30°
(c) 60°
(d) 90°

(d) 90o

#### Question 26:

If (sin θ − cos θ) = 0, then (sin4 θ + cos4 θ) = ?
(a) 1
(b) $\frac{1}{2}$
(c) $\frac{1}{4}$
(d) $\frac{3}{4}$

(b) $\frac{1}{2}$

#### Question 27:

If ∆ABC is right-angled at C, then cos (A + B) = ?
(a) 0
(b) $\frac{1}{2}$
(c) 1
(d) $\frac{\sqrt{3}}{2}$

(a) 0

A + ∠B + ∠C = 180o   (by angle sum property)
It is given that ∠C = 90o.
∴ ∠A + ∠B = 90o
cos (A + B) = cos (90o) = 0

#### Question 28:

A pole 6 m high casts a shadow $2\sqrt{3}$ m long on the ground. The sun's altitude is
(a) 30°
(b) 45°
(c) 60°
(d) 90°