RS Aggarwal 2015 Solutions for Class 10 Math Chapter 5 - Trigonometric Ratios

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Page No 273:

Question 1:

If sin $θ = \frac{\sqrt{3}}{2}$ , find the value of all T-ratios of θ.

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that sin $θ$ = $\frac{perpendicular}{hypotenuse}$ = $\frac{A B}{A C}$ = $\frac{\sqrt{3}}{2}$ .

So, if AB = $\sqrt{3} k$ , then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ BC² = AC² $-$ AB² = (2k)² $-$ ( $\sqrt{3} k$ )²
⇒ BC² = 4k² $-$ 3k² = k²
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
cos $θ$ = $\frac{B C}{A C}$ = $\frac{k}{2 k} = \frac{1}{2}$
tan $θ$ = $\frac{A B}{B C} = \frac{\sqrt{3} k}{k} = \sqrt{3}$

∴ cot $θ$ = $\frac{1}{\tan θ} = \frac{1}{\sqrt{3}}$ , cosec $θ$ = $\frac{1}{\sin θ} = \frac{2}{\sqrt{3}}$ and sec $θ$ = $\frac{1}{\cos θ} = 2$

Page No 273:

Question 2:

If cos $θ = \frac{7}{25}$ find the values of all T-ratios of θ.

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that cos $θ$ = $\frac{Base}{hypotenuse}$ = $\frac{B C}{A C}$ = $\frac{7}{25}$ .

So, if BC = 7k, then AC = 25k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC² = (25k)² $-$ (7k)².
⇒ AB² = 625k² $-$ 49k² = 576k²
⇒ AB = 24k
Now, finding the other trigonometric ratios using their definitions, we get:
sin $θ$ = $\frac{A B}{A C}$ = $\frac{24 k}{25 k} = \frac{24}{25}$
tan $θ$ = $\frac{A B}{B C} = \frac{24 k}{7 k} = \frac{24}{7}$
∴ cot $θ$ = $\frac{1}{\tan θ} = \frac{7}{24}$ , cosec $θ$ = $\frac{1}{\sin θ} = \frac{25}{24}$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{25}{7}$

Page No 273:

Question 3:

If tan $θ = \frac{15}{8}$ find the values of all T-ratios of θ.

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that tan $θ$ = $\frac{Perpendicular}{Base}$ = $\frac{A B}{B C}$ = $\frac{15}{8}$ .

So, if BC = 8k, then AB = 15k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC² = (15k)² + (8k)²
⇒ AC² = 225k² + 64k² = 289k²
⇒ AC = 17k

Now, finding the other T-ratios using their definitions, we get:
sin $θ$ = $\frac{A B}{A C}$ = $\frac{15 k}{17 k} = \frac{15}{17}$
cos $θ$ = $\frac{B C}{A C} = \frac{8 k}{17 k} = \frac{8}{17}$

∴ cot $θ$ = $\frac{1}{\tan θ} = \frac{8}{15}$ , cosec $θ$ = $\frac{1}{\sin θ} = \frac{17}{15}$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{17}{8}$

Page No 273:

Question 4:

If cot θ = 2, find the value of all T-ratios of θ.

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that cot $θ$ = $\frac{base}{Perpendicular}$ = $\frac{B C}{A B}$ = 2.

So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC² = (2k)² + (k)²
⇒ AC² = 4k² + k² = 5k²
⇒ AC = $\sqrt{5}$ k
Now, finding the other T-ratios using their definitions, we get:
sin $θ$ = $\frac{A B}{A C}$ = $\frac{k}{\sqrt{5} k} = \frac{1}{\sqrt{5}}$
cos $θ$ = $\frac{B C}{A C} = \frac{2 k}{\sqrt{5} k} = \frac{2}{\sqrt{5}}$

∴ tan $θ$ = $\frac{1}{c o t θ} = \frac{1}{2}$ , cosec $θ$ = $\frac{1}{\sin θ} = \sqrt{5}$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{\sqrt{5}}{2}$

Page No 273:

Question 5:

If cosec θ = $\sqrt{10}$ , the find the values of all T-ratios of θ.

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that cosec $θ$ = $\frac{Hypotenuse}{Perpendicular}$ = $\frac{A C}{A B}$ = $\frac{\sqrt{10}}{1}$ .

So, if AC = ( $\sqrt{10}$ )k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ BC² = AC² $-$ AB² = 10k² $-$ k²
⇒ BC² = 9k²
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
tan $θ$ = $\frac{A B}{B C}$ = $\frac{k}{3 k} = \frac{1}{3}$

cos $θ$ = $\frac{B C}{A C} = \frac{3 k}{\sqrt{10} k} = \frac{3}{\sqrt{10}}$

∴ $\sin θ = \frac{1}{\cos e c θ} = \frac{1}{\sqrt{10}}$ , cot $θ$ = $\frac{1}{\tan θ} = 3$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{\sqrt{10}}{3}$

Page No 273:

Question 6:

If cot θ = $\frac{15}{8}$ , show then evaluate $\frac{(2 + 2 sinθ) (1 - sinθ)}{(1 + cosθ) (2 - 2 cosθ)} .$

Answer:

Let us consider a right $△$ ABC, right angled at B.
Now, we know that cot $θ$ = $\frac{Base}{Perpendicular}$ = $\frac{B C}{A B}$ = $\frac{15}{8}$ .

Using Pythagoras theorem, we have:
AC² = AB² + BC² = (8k)² + (15k)²
⇒ AC² = 64k²+ 225k² = 289k²
⇒ AC = 17k

Finding out the value of sin $θ$ and cos $θ$ using their definitions, we have:
sin $θ$ = $\frac{A B}{A C} = \frac{8 k}{17 k} = \frac{8}{17}$
cos $θ$ = $\frac{B C}{A C} = \frac{15 k}{17 k} = \frac{15}{17}$

Substituting these values in the given expression, we get:
$\frac{(2 + 2 s i n θ) (1 - s i n θ)}{(1 + c o s θ) (2 - 2 c o s θ)} = \frac{(2 + 2 \times \frac{8}{17}) (1 - \frac{8}{17})}{(1 + \frac{15}{17}) (2 - 2 \times \frac{15}{17})} = \frac{(2 + \frac{16}{17}) (1 - \frac{8}{17})}{(1 + \frac{15}{17}) (2 - \frac{30}{17})} = \frac{(\frac{34 + 16}{17}) (\frac{9}{17})}{(\frac{32}{17}) (\frac{34 - 30}{17})} = \frac{\frac{50}{17} \times \frac{9}{17}}{\frac{32}{17} \times \frac{4}{17}} = \frac{225}{64}$

Page No 273:

Question 7:

If tan θ = $\frac{4}{3}$ , show that (sin θ + cos θ) = $\frac{7}{5}$ .

Answer:

Let us consider a right $△$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that tan $θ$ = $\frac{A B}{B C}$ = $\frac{4}{3}$ .

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC² = (4k)² + (3k)²
⇒ AC² = 16k²+ 9k² = 25k²
⇒ AC = 5k
Finding out the values of sin $θ$ and cos $θ$ using their definitions, we have:
sin $θ$ = $\frac{A B}{A C} = \frac{4 k}{5 k} = \frac{4}{5}$
cos $θ$ = $\frac{B C}{A C} = \frac{3 k}{5 k} = \frac{3}{5}$
Substituting these values in the given expression, we get:
(sin $θ$ + cos $θ$ ) = $(\frac{4}{5} + \frac{3}{5}) = (\frac{7}{5}) = RHS$
i.e., LHS = RHS

Hence proved.

Page No 273:

Question 8:

If tan θ = $\frac{1}{\sqrt{7}}$ , show that $\frac{({cosec}^{2} θ - \sec^{2} θ)}{({cosec}^{2} θ + \sec^{2} θ)} = \frac{3}{4} .$

Answer:

Let us consider a right $△$ ABC, right angled at B and $∠ C = θ$ .
Now it is given that tan $θ$ = $\frac{A B}{B C}$ = $\frac{1}{\sqrt{7}}$ .

So, if AB = k, then BC = $\sqrt{7}$ k, where k is a positive number.
Using Pythagoras theorem, we have:
AC²= AB² + BC²
⇒ AC² = (k)² + ( $\sqrt{7}$ k)²
⇒ AC² = k² + 7k²
⇒ AC = 2 $\sqrt{2}$ k
Now, finding out the values of the other trigonometric ratios, we have:
sin $θ$ = $\frac{A B}{A C} = \frac{k}{2 \sqrt{2} k} = \frac{1}{2 \sqrt{2}}$
cos $θ$ = $\frac{B C}{A C} = \frac{\sqrt{7} k}{2 \sqrt{2} k} = \frac{\sqrt{7}}{2 \sqrt{2}}$
∴ cosec $θ$ = $\frac{1}{\sin θ} = 2 \sqrt{2}$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{2 \sqrt{2}}{\sqrt{7}}$
Substituting the values of cosec $θ$ and sec $θ$ in the given expression, we get:
$\frac{\cos e c^{2} θ - s e c^{2} θ}{\cos e c^{2} θ + s e c^{2} θ} = \frac{(2 \sqrt{2})^{2} - {(\frac{2 \sqrt{2}}{\sqrt{7}})}^{2}}{(2 \sqrt{2})^{2} + {(\frac{2 \sqrt{2}}{\sqrt{7}})}^{2}} = \frac{8 - (\frac{8}{7})}{8 + (\frac{8}{7})} = \frac{\frac{56 - 8}{7}}{\frac{56 + 8}{7}} = \frac{48}{64} = \frac{3}{4} = RHS$
i.e., LHS = RHS

Hence proved.

Page No 273:

Question 9:

If cosec θ = 2, show that $(cotθ + \frac{sinθ}{1 + cosθ}) = 2 .$

Answer:

Let us consider a right $△$ ABC, right angled at B and $∠ C = θ$ .
Now, it is given that cosec $θ$ = 2.
Also, sin $θ$ = $\frac{1}{\cos e c θ} = \frac{1}{2} = \frac{A B}{A C}$

So, if AB = k, then AC = 2k, where k is a positive number.
Using Pythagoras theorem, we have:
⇒ AC² = AB² + BC²
⇒ BC² = AC² $-$ AB²
⇒ BC² = (2k)² $-$ (k)²
⇒ BC² = 3k²
⇒ BC = $\sqrt{3}$ k
Finding out the other T-ratios using their definitions, we get:
cos $θ$ = $\frac{B C}{A C} = \frac{\sqrt{3} k}{2 k} = \frac{\sqrt{3}}{2}$
tan $θ$ = $\frac{A B}{B C} = \frac{k}{\sqrt{3} k} = \frac{1}{\sqrt{3}}$
cot $θ$ = $\frac{1}{\tan θ} = \sqrt{3}$
Substituting these values in the given expression, we get:
$c o t θ + \frac{\sin θ}{1 + \cos θ} = \sqrt{3} + \frac{(\frac{1}{2})}{1 + \frac{\sqrt{3}}{2}} = \sqrt{3} + \frac{\frac{1}{2}}{\frac{2 + \sqrt{3}}{2}}$

$= \sqrt{3} + \frac{1}{2 + \sqrt{3}} = \frac{\sqrt{3} (2 + \sqrt{3}) + 1}{2 + \sqrt{3}} = \frac{2 \sqrt{3} + 3 + 1}{2 + \sqrt{3}} = \frac{2 (2 + \sqrt{3})}{2 + \sqrt{3}} = 2$
i.e., LHS = RHS

Hence proved.

Page No 273:

Question 10:

If sec θ = $\frac{5}{4}$ , verify that $\frac{tanθ}{(1 + \tan^{2} θ)} = \frac{sinθ}{s e c θ}$ .

Answer:

Let us consider a right $△$ ABC, right angled at B and $∠ C = θ$ .
Now, it is given that sec $θ$ = $\frac{5}{4}$ .
Also, cos $θ$ = $\frac{1}{s e c θ} = \frac{4}{5} = \frac{B C}{A C}$

So, if BC = 4k, then AC = 5k, where k is a positive number.
Using Pythagoras theorem, we have:
⇒ AC² = AB² + BC²
⇒ AB² = AC² $-$ BC²
⇒ AB² = (5k)² $-$ (4k)² = 25k² $-$ 16k²
⇒ AB² = 9k²
⇒ AB = 3k

Finding out the other T-ratios using their definitions, we get:
sin $θ$ = $\frac{A B}{A C} = \frac{3 k}{5 k} = \frac{3}{5}$

tan $θ$ = $\frac{A B}{B C} = \frac{3 k}{4 k} = \frac{3}{4}$

Substituting the values in LHS and RHS, we get:
$LHS = \frac{\tan θ}{1 + \tan^{2} θ} = \frac{\frac{3}{4}}{1 + {(\frac{3}{4})}^{2}} = \frac{\frac{3}{4}}{1 + \frac{9}{16}} = \frac{\frac{3}{4}}{\frac{16 + 9}{16}} = \frac{12}{25} R HS = \frac{\sin θ}{s e c θ} = \frac{\frac{3}{5}}{\frac{5}{4}} = \frac{12}{25}$
i.e., LHS = RHS

Hence proved.

Page No 273:

Question 11:

If cos θ = 0.6, show that (5 sin θ − 3 tan θ) = 0.

Answer:

Let us consider a right $△$ ABC right angled at B.
Now, we know that cos $θ$ = 0.6 = $\frac{B C}{A C}$ = $\frac{3}{5}$

So, if BC = 3k, then AC = 5k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC²
⇒ AB² = (5k)² $-$ (3k)² = 25k² $-$ 9k²
⇒ AB² = 16k²
⇒ AB = 4k

Finding out the other T-ratios using their definitions, we get:
sin $θ$ = $\frac{A B}{A C} = \frac{4 k}{5 k} = \frac{4}{5}$

tan $θ$ = $\frac{A B}{B C} = \frac{4 k}{3 k} = \frac{4}{3}$

Substituting the values in the given expression, we get:
5 sin $θ$ $-$ 3 tan $θ$
$\Rightarrow 5 (\frac{4}{5}) - 3 (\frac{4}{3}) \Rightarrow 4 - 4 = 0 = RHS$
i.e., LHS = RHS

Hence proved.

Page No 273:

Question 12:

If 3 tan θ = 4, show that $\frac{4 cosθ - sinθ}{2 cosθ + sinθ} = \frac{4}{5} .$

Answer:

Let us consider a right $△$ ABC right angled at B and $∠ C = θ$ .
We know that tan $θ$ = $\frac{A B}{B C}$ = $\frac{4}{3}$

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AC² = 16k² + 9k²
⇒ AC² = 25k²
⇒ AC = 5k

Now, we have:

sin $θ$ = $\frac{A B}{A C} = \frac{4 k}{5 k} = \frac{4}{5}$

cos $θ$ = $\frac{B C}{A C} = \frac{3 k}{5 k} = \frac{3}{5}$

Substituting these values in the given expression, we get:

$\frac{4 \cos θ - \sin θ}{2 \cos θ + \sin θ} = \frac{4 (\frac{3}{5}) - \frac{4}{5}}{2 (\frac{3}{5}) + \frac{4}{5}} = \frac{\frac{12}{5} - \frac{4}{5}}{\frac{6}{5} + \frac{4}{5}} = \frac{\frac{12 - 4}{5}}{\frac{6 + 4}{5}} = \frac{8}{10} = \frac{4}{5} = RHS$
i.e., LHS = RHS

Hence proved.

Page No 273:

Question 13:

If tan θ = $\frac{a}{b}$ , show that $(\frac{a sinθ - b cosθ}{a sinθ + b cosθ}) = \frac{(a^{2} - b^{2})}{(a^{2} + b^{2})} .$

Answer:

It is given that tan $θ = \frac{a}{b}$ .

LHS = $\frac{a \sin θ - b \cos θ}{a \sin θ + b \cos θ}$
Dividing the numerator and denominator by cos $θ$ , we get:

$\frac{a \tan θ - b}{a \tan θ + b}$ (∵ tan $θ = \frac{\sin θ}{\cos θ}$ )
Now, substituting the value of tan $θ$ in the above expression, we get:
$\frac{a (\frac{a}{b}) - b}{a (\frac{a}{b}) + b} = \frac{\frac{a^{2}}{b} - b}{\frac{a^{2}}{b} + b} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}} = RHS$
i.e., LHS = RHS

Hence proved.

Page No 273:

Question 14:

If 3 cot θ = 2, show that $(\frac{4 sinθ - 3 cosθ}{2 sinθ + 6 cosθ}) = \frac{1}{3} .$

Answer:

It is given that cot $θ = \frac{2}{3}$ .

LHS = $\frac{4 \sin θ - 3 \cos θ}{2 \sin θ + 6 \cos θ}$
Dividing the above expression by sin $θ$ , we get:
$\frac{4 - 3 c o t θ}{2 + 6 c o t θ}$ [∵ cot $θ = \frac{\cos θ}{\sin θ}$ ]
Now, substituting the values of cot $θ$ in the above expression, we get:
$\frac{4 - 3 (\frac{2}{3})}{2 + 6 (\frac{2}{3})} = \frac{4 - 2}{2 + 4} = \frac{2}{6} = \frac{1}{3}$
i.e., LHS = RHS

Hence proved.

Page No 273:

Question 15:

If sec θ = $\frac{17}{8}$ , verify that $\frac{3 - 4 \sin^{2} θ}{4 \cos^{2} θ - 3} = \frac{3 - \tan^{2} θ}{1 - 3 \tan^{2} θ}$ .

Answer:

It is given that sec $θ$ = $\frac{17}{8}$ .

Let us consider a right $△$ ABC right angled at B and $∠ C = θ$ .
We know that cos $θ$ = $\frac{1}{s e c θ} = \frac{8}{17} = \frac{B C}{A C}$

So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC² = (17k)² $-$ (8k)²
⇒ AB² = 289k² $-$ 64k² = 225k²
⇒ AB = 15k.

Now, tan $θ$ = $\frac{A B}{B C} = \frac{15}{8}$ and sin $θ$ = $\frac{A B}{A C} = \frac{15 k}{17 k} = \frac{15}{17}$

The given expression is $\frac{3 - 4 \sin^{2} θ}{4 \cos^{2} θ - 3} = \frac{3 - \tan^{2} θ}{1 - 3 \tan^{2} θ}$ .

Substituting the values in the above expression, we get:
$LHS = \frac{3 - 4 {(\frac{15}{17})}^{2}}{4 {(\frac{8}{17})}^{2} - 3} = \frac{3 - \frac{900}{289}}{\frac{256}{289} - 3} = \frac{867 - 900}{256 - 867} = \frac{- 33}{- 611} = \frac{33}{611}$

$RHS = \frac{3 - {(\frac{15}{8})}^{2}}{1 - 3 {(\frac{15}{8})}^{2}} = \frac{3 - \frac{225}{64}}{1 - \frac{675}{64}} = \frac{192 - 225}{64 - 675} = \frac{- 33}{- 611} = \frac{33}{611}$

∴ LHS = RHS
Hence proved.

Page No 273:

Question 16:

If tan θ = $\frac{20}{21}$ , show that $\frac{(1 - sinθ + cosθ)}{(1 + sinθ + cosθ)} = \frac{3}{7} .$

Answer:

Let us consider a right $△$ ABC right angled at B and $∠ C = θ$ .
Now, we know that tan $θ$ = $\frac{A B}{B C}$ = $\frac{20}{21}$

So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC²= (20k)²+ (21k)²
⇒ AC² = 841k²
⇒ AC = 29k
Now, sin $θ$ = $\frac{A B}{A C} = \frac{20}{29}$ and cos $θ$ = $\frac{B C}{A C} = \frac{21}{29}$

Substituting these values in the given expression, we get:
$LHS = \frac{1 - \sin θ + \cos θ}{1 + \sin θ + \cos θ} = \frac{1 - \frac{20}{29} + \frac{21}{29}}{1 + \frac{20}{29} + \frac{21}{29}} = \frac{\frac{29 - 20 + 21}{29}}{\frac{29 + 20 + 21}{29}} = \frac{30}{70} = \frac{3}{7} = RHS$
∴ LHS = RHS

Hence proved.

Page No 273:

Question 17:

In a ∆ABC, right-angled at B, it is given that AB = 12 cm and BC = 5 cm.

Find the value of
(i) cos A
(ii) cosec A
(iii) cos C
(iv) coses C.

Answer:

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC² = 12² + 5² = 144 + 25
⇒ AC² = 169
⇒ AC = 13 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i) cos A = $\frac{A B}{A C} = \frac{12}{13}$
(ii) cosec A = $\frac{1}{\sin A} = \frac{A C}{B C} = \frac{13}{5}$

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) cos C = $\frac{B C}{A C} = \frac{5}{13}$
(iv) cosec C = $\frac{1}{\sin C} = \frac{A C}{A B} = \frac{13}{12}$

Page No 274:

Question 18:

In a ∆ABC, it is given that ∠B = 90°, AB = 24 cm and BC = 7 cm.
Figure

Find the value of
(i) sin A
(ii) cos A
(iii) sin C
(iv) cos C

Answer:

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC²= (24)² + (7)²
⇒ AC² = 576 + 49 = 625
⇒ AC = 25 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i) sin A = $\frac{B C}{A C} = \frac{7}{25}$

(ii) cos A = $\frac{A B}{A C} = \frac{24}{25}$

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) sin C = $\frac{A B}{A C} = \frac{24}{25}$

(iv) cos C = $\frac{B C}{A C} = \frac{7}{25}$

Page No 274:

Question 19:

Given a ∆ABC, in which ∠C = 90°, ∠ABC = θ°, BC = 21 units, AB = 29 units.
Show that (cos²θ − sin²θ) = $\frac{41}{841} .$

Answer:

Using Pythagoras theorem, we get:
AB²=AC²+ BC²
⇒ AC² = AB² $-$ BC²
⇒ AC² = (29)² $-$ (21)²
⇒ AC² = 841 $-$ 441
⇒ AC²= 400
⇒ AC = $\sqrt{400}$ = 20 units

Now, sin $θ = \frac{A C}{A B} = \frac{20}{29}$ and cos $θ$ = $\frac{B C}{A B} = \frac{21}{29}$

cos² $θ$ $-$ sin² $θ$ = ${(\frac{21}{29})}^{2} - {(\frac{20}{29})}^{2} = \frac{441}{841} - \frac{400}{841} = \frac{41}{841}$

Hence Proved.

Page No 276:

Question 1:

If cosθ = $\frac{4}{5}$ , then tanθ = ?
(a) $\frac{3}{4}$
(b) $\frac{4}{3}$
(c) $\frac{3}{5}$
(d) $\frac{5}{3}$

Answer:

(a) $\frac{3}{4}$
Since cosθ = $\frac{4}{5}$ but cosθ = $\frac{A B}{A C}$
So, $\frac{A B}{A C}$ = $\frac{4}{5}$
Thus, AB = 4k and AC = 5k

Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
⇒ BC² = AC² $-$ AB²
⇒ BC²= (5k)² $-$ (4k)²
⇒ BC² = 9k²
⇒ BC = 3k
∴ tanθ = $\frac{B C}{A B}$ = $\frac{3}{4}$

Page No 276:

Question 2:

If sinθ = $\frac{1}{2}$ , then cotθ = ?
(a) $\frac{1}{\sqrt{3}}$
(b) $\sqrt{3}$
(c) $\frac{\sqrt{3}}{2}$
(d) 1

Answer:

(b) $\sqrt{3}$

Given: sinθ = $\frac{1}{2}$ , but sinθ = $\frac{BC}{AC}$
So, $\frac{BC}{AC}$ = $\frac{1}{2}$
Thus, BC = k and AC = 2k

Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
AB² = AC² $-$ BC²
AB²= (2k)² $-$ (k)²
AB² = 3k²
AB = $\sqrt{3}$ k
So, tanθ = $\frac{BC}{AB}$ = $\frac{k}{\sqrt{3} k}$ = $\frac{1}{\sqrt{3}}$
∴ cotθ = $\frac{1}{\tan θ}$ = $\sqrt{3}$

Page No 276:

Question 3:

If tan θ = $\frac{3}{4}$ , then cosec θ = ?
(a) $\frac{4}{3}$
(b) $\frac{4}{5}$
(c) $\frac{5}{4}$
(d) $\frac{5}{3}$

Answer:

(d) $\frac{5}{3}$

Given: tan $θ$ = $\frac{3}{4}$ , but tan $θ$ = $\frac{Perpendicular}{Base}$
So, let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
So, $\frac{B C}{A B}$ = $\frac{3}{4}$
Thus, BC = 3k and AB = 4k

Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
⇒ AC² = (4k)²+ (3k)²
⇒ AC² = 25k²
⇒ AC = 5k
∴ cosec $θ$ = $\frac{A C}{B C}$ = $\frac{5 k}{3 k}$ = $\frac{5}{3}$

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Question 4:

If cosec θ = $\sqrt{10}$ , then sec θ = ?
(a) $\frac{3}{\sqrt{10}}$
(b) $\frac{\sqrt{10}}{3}$
(c) $\frac{1}{\sqrt{10}}$
(d) $\frac{2}{\sqrt{10}}$

Answer:

(b) $\frac{\sqrt{10}}{3}$
Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
Given: cosec $θ$ = $\sqrt{10}$ , but sin $θ$ = $\frac{1}{\cos e c θ}$ = $\frac{1}{\sqrt{10}}$
Also, sin $θ$ = $\frac{Perpendicular}{Hypotenuse}$ = $\frac{B C}{A C}$
So, $\frac{B C}{A C}$ = $\frac{1}{\sqrt{10}}$
Thus, BC = k and AC = $\sqrt{10}$ k

Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC²
⇒ AB²= ( $\sqrt{10}$ k)² $-$ (k)²
⇒ AB² = 9k²
⇒ AB = 3k
∴ sec $θ$ = $\frac{A C}{A B}$ = $\frac{\sqrt{10} k}{3 k} = \frac{\sqrt{10}}{3}$

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Question 5:

If tan θ = $\frac{8}{15}$ , then cosec θ = ?
(a) $\frac{17}{8}$
(b) $\frac{8}{17}$
(c) $\frac{17}{15}$
(d) $\frac{15}{17}$

Answer:

(a) $\frac{17}{8}$

Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
Given: tan $θ$ = $\frac{8}{15}$ , but tan $θ$ = $\frac{B C}{A B}$
So, $\frac{B C}{A B}$ = $\frac{8}{15}$
Thus, BC = 8k and AB = 15k

Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
⇒ AC² = (15k)²+ (8k)²
⇒ AC² = 289k²
⇒ AC = 17k
∴ cosec $θ$ = $\frac{A C}{B C} = \frac{17 k}{8 k} = \frac{17}{8}$

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Question 6:

If sin θ $\frac{a}{b}$ , then cos θ = ?
(a) $\frac{b}{\sqrt{b^{2} - a^{2}}}$
(b) $\frac{\sqrt{b^{2} - a^{2}}}{b}$
(c) $\frac{a}{\sqrt{b^{2} - a^{2}}}$
(d) $\frac{b}{a}$

Answer:

(b) $\frac{\sqrt{b^{2} - a^{2}}}{b}$
Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
Given: sin θ = $\frac{a}{b}$ , but sin θ = $\frac{B C}{A C}$
So, $\frac{B C}{A C}$ = $\frac{a}{b}$
Thus, BC = ak and AC = bk

Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC²
⇒ AB²= (bk)² $-$ (ak)²
⇒ AB² = $(b^{2} - a^{2})$ k²
⇒ AB = ( $\sqrt{b^{2} - a^{2}}$ )k
∴ cos θ = $\frac{AB}{AC}$ = $\frac{\sqrt{b^{2} - a^{2}} k}{b k}$ = $\frac{\sqrt{b^{2} - a^{2}}}{b}$

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Question 7:

If tan θ = $\sqrt{3}$ , then sec θ = ?
(a) $\frac{2}{\sqrt{3}}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{1}{2}$
(d) 2

Answer:

(d) 2

Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
Given: tan θ = $\sqrt{3}$
But tan θ = $\frac{B C}{A B}$
So, $\frac{B C}{A B}$ = $\frac{\sqrt{3}}{1}$
Thus, BC = $\sqrt{3}$ k and AB = k

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC² = ( $\sqrt{3}$ k)² + (k)²
⇒ AC²= 4k²
⇒ AC = 2k
∴ sec θ = $\frac{AC}{AB} = \frac{2 k}{k} = \frac{2}{1}$

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Question 8:

If sec θ = $\frac{25}{7}$ , then sin θ = ?
(a) $\frac{7}{24}$
(b) $\frac{24}{7}$
(c) $\frac{24}{25}$
(d) None of these

Answer:

(c) $\frac{24}{25}$
Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
Given: sec θ = $\frac{25}{7}$
But cos θ = $\frac{1}{s e c θ}$ = $\frac{A B}{A C}$ = $\frac{7}{25}$
Thus, AC = 25k and AB = 7k

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ BC² = AC² $-$ AB²
⇒ BC² = (25k)² $-$ (7k)²
⇒ BC²= 576k²
⇒ BC = 24k

∴ sin θ = $\frac{B C}{A C} = \frac{24 k}{25 k} = \frac{24}{25}$

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Question 9:

If sin θ = $\frac{\sqrt{3}}{2}$ , then (cosecθ + cotθ) = ?
(a) $2 \sqrt{3}$
(b) $\frac{2 \sqrt{3}}{3}$
(c) $(2 + \sqrt{3})$
(d) $\sqrt{3}$

Answer:

(d) $\sqrt{3}$
Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
sin θ = $\frac{\sqrt{3}}{2}$ = $\frac{B C}{A C}$
So, BC = $\sqrt{3}$ k and AC = 2k

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC²
⇒ AB² = (2k)² $-$ ( $\sqrt{3}$ k)²
⇒ AB² = k²
⇒ AB = k

So, cot θ = $\frac{A B}{B C}$ = $\frac{k}{\sqrt{3} k} = \frac{1}{\sqrt{3}}$
cosec θ = $\frac{1}{\sin θ} = \frac{2}{\sqrt{3}}$
∴ (cosec θ + cot θ) = $\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}}$ = $\frac{3}{\sqrt{3}} = \sqrt{3}$

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Question 10:

If tan θ = $\frac{4}{3}$ , then (sinθ + cosθ) = ?
(a) $\frac{7}{3}$
(b) $\frac{7}{4}$
(c) $\frac{7}{5}$
(d) $\frac{5}{7}$

Answer:

(c) $\frac{7}{5}$
Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
tan θ = $\frac{4}{3} = \frac{B C}{A B}$
So, AB = 3k and BC = 4k

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC² = (3k)² + (4k)²
⇒ AC² = 25k²
⇒ AC= 5k
Thus, sin θ = $\frac{B C}{A C}$ = $\frac{4}{5}$
and cos θ = $\frac{A B}{A C} = \frac{3}{5}$
∴ (sin θ + cos θ) = ( $\frac{4}{5}$ + $\frac{3}{5}$ ) = $\frac{7}{5}$

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Question 11:

If 3x = cosec θ and $\frac{3}{x}$ = cot θ, than $3 (x^{2} - \frac{1}{x^{2}}) = ?$
(a) $\frac{1}{27}$
(b) $\frac{1}{81}$
(c) $\frac{1}{3}$
(d) $\frac{1}{9}$

Answer:

(c) $\frac{1}{3}$
Given: 3x = cosec θ and $\frac{3}{x}$ = cot θ
Also, we can deduce that x = $\frac{\cos e c θ}{3}$ and $\frac{1}{x} = \frac{c o t θ}{3}$ .
So, substituting the values of x and $\frac{1}{x}$ in the given expression, we get:
3 $(x^{2} - \frac{1}{x^{2}})$ = 3 $({(\frac{\cos e c θ}{3})}^{2} - {(\frac{c o t θ}{3})}^{2})$
= 3 $((\frac{\cos e c^{2} θ}{9}) - (\frac{c o t^{2} θ}{9}))$
= $\frac{3}{9} (\cos e c^{2} θ - c o t^{2} θ)$
= $\frac{1}{3}$ [By using the identity: $(\cos e c^{2} θ - c o t^{2} θ = 1)$ ]

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Question 12:

If 2x = sec A and $\frac{2}{x}$ = tan A, then $2 (x^{2} - \frac{1}{x^{2}}) = ?$ =?
(a) $\frac{1}{2}$
(b) $\frac{1}{4}$
(c) $\frac{1}{8}$
(d) $\frac{1}{16}$

Answer:

(a) $\frac{1}{2}$
Given: 2x = sec A and $\frac{2}{x}$ = tan A
Also, we can deduce that x = $\frac{\sec A}{2}$ and $\frac{1}{x} = \frac{\tan A}{2}$ .
So, substituting the values of x and $\frac{1}{x}$ in the given expression, we get:
2 $(x^{2} - \frac{1}{x^{2}})$ = 2 $({(\frac{\sec A}{2})}^{2} - {(\frac{\tan A}{2})}^{2})$
= 2 $((\frac{\sec^{2} A}{4}) - (\frac{\tan^{2} A}{4}))$
= $\frac{2}{4} (\sec^{2} A - \tan^{2} A)$
= $\frac{1}{2}$ [By using the identity: $(\sec^{2} θ - \tan^{2} θ = 1)$ ]

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Question 13:

If (tan θ + cot θ) = 5, then (tan² θ + cot² θ) = ?

(a) 27
(b) 25
(c) 24
(d) 23

Answer:

(d) 23
We have (tan θ +cot θ) = 5
Squaring both sides, we get:
(tan θ +cot θ)² = 5²
⇒ tan² θ + cot² θ + 2 tan θ cot θ = 25
⇒ tan² θ + cot² θ + 2 = 25 [∵ tan θ = $\frac{1}{\cot θ}$ ]
⇒ tan² θ + cot² θ = 25 − 2 = 23

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Question 14:

If (cos θ + sec θ) = $\frac{5}{2}$ , then (cos² θ + sec² θ) = ?
(a) $\frac{21}{4}$
(b) $\frac{17}{4}$
(c) $\frac{29}{4}$
(d) $\frac{33}{4}$

Answer:

(b) $\frac{17}{4}$
We have (cos θ +sec θ) = $\frac{5}{2}$
Squaring both sides, we get:
(cos θ + sec θ)² = ( $\frac{5}{2}$ )²
$\Rightarrow$ cos² θ + sec² θ + 2 cos θ sec θ = $\frac{25}{4}$
$\Rightarrow$ cos² θ + sec² θ + 2 = $\frac{25}{4}$ [∵ sec θ = $\frac{1}{\cos θ}$ ]
$\Rightarrow$ cos² θ + sec² θ = $\frac{25}{4}$ − 2 = $\frac{17}{4}$

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Question 15:

If tan θ = $\frac{1}{\sqrt{7}}$ , then $\frac{(\cos e c^{2} θ - s e c^{2} θ)}{(\cos e c^{2} θ + s e c^{2} θ)} = ?$ = ?
(a) $\frac{- 2}{3}$
(b) $\frac{- 3}{4}$
(c) $\frac{2}{3}$
(d) $\frac{3}{4}$

Answer:

(d) $\frac{3}{4}$
$= \frac{{cosec}^{2} θ - \sec^{2} θ}{{cosec}^{2} θ + \sec^{2} θ} = \frac{\sin^{2} θ (\frac{1}{\sin^{2} θ} - \frac{1}{\cos^{2} θ})}{\sin^{2} θ (\frac{1}{\sin^{2} θ} + \frac{1}{\cos^{2} θ})} [Multiplying the numerator and denominator by \sin^{2} θ] = \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}$
= $\frac{1 - \frac{1}{7}}{1 + \frac{1}{7}} = \frac{6}{8} = \frac{3}{4}$

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Question 16:

If 7 tan θ = 4, then $\frac{(7 sinθ - 3 cosθ)}{(7 sinθ + 3 cosθ)}$ = ?
(a) $\frac{1}{7}$
(b) $\frac{5}{7}$
(c) $\frac{3}{7}$
(d) $\frac{5}{14}$

Answer:

(a) $\frac{1}{7}$

7 tan θ = 4

Now, dividing the numerator and denominator of the given expression by cos θ, we get:
$\frac{\frac{1}{\cos θ} (7 \sin θ - 3 \cos θ)}{\frac{1}{\cos θ} (7 \sin θ + 3 \cos θ)}$

= $\frac{7 \tan θ - 3}{7 \tan θ + 3}$

= $\frac{4 - 3}{4 + 3}$ [∵ 7 tan θ = 4]
= $\frac{1}{7}$

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Question 17:

If 3 cot θ = 4, then $\frac{(5 sinθ + 3 cosθ)}{(5 sinθ - 3 cosθ)}$ = ?
(a) $\frac{1}{3}$
(b) 3
(c) $\frac{1}{9}$
(d) 9

Answer:

(d) 9

We have $\frac{(5 \sin θ + 3 \cos θ)}{(5 \sin θ - 3 \cos θ)}$ .

Dividing the numerator and denominator of the given expression by sin θ, we get:
$\frac{\frac{1}{\sin θ} (5 \sin θ + 3 \cos θ)}{\frac{1}{\sin θ} (5 \sin θ - 3 \cos θ)}$

= $\frac{5 + 3 c o t θ}{5 - 3 co t θ}$

= $\frac{5 + 4}{5 - 4}$ = 9 [∵ 3 cot θ = 4]

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Question 18:

If θ = $\frac{a}{b}$ , then $\frac{(a sinθ - b cosθ)}{(a \sin θ + b cosθ)}$ = ?
(a) $\frac{(a^{2} + b^{2})}{(a^{2} - b^{2})}$
(b) $\frac{(a^{2} - b^{2})}{(a^{2} + b^{2})}$
(c) $\frac{a^{2}}{(a^{2} + b^{2})}$
(d) $\frac{b^{2}}{(a^{2} + b^{2})}$

Answer:

(b) $\frac{(a^{2} - b^{2})}{(a^{2} + b^{2})}$
We have tan θ = $\frac{a}{b}$

Now, dividing the numerator and denominator of the given expression by cos θ, we get:
$\frac{a \sin θ - b \cos θ}{a \sin θ + b \cos θ} = \frac{\frac{1}{\cos θ} (a \sin θ - b \cos θ)}{\frac{1}{\cos θ} (a \sin θ + b \cos θ)} = \frac{a \tan θ - b}{a \tan θ + b} = \frac{\frac{a^{2}}{b} - b}{\frac{a^{2}}{b} + b} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$

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Question 19:

If cot θ = , then $\frac{(2 + 2 sinθ) (1 - sinθ)}{(1 + cosθ) (2 - 2 cosθ)}$ = ?
(a) $\frac{64}{225}$
(b) $\frac{225}{64}$
(c) $\frac{64}{289}$
(d) $\frac{289}{64}$

Answer:

(b) $\frac{225}{64}$
We have:

$\frac{(2 + 2 \sin θ) (1 - \sin θ)}{(1 + \cos θ) (2 - 2 \cos θ)} = \frac{2 (1 + \sin θ) (1 - \sin θ)}{2 (1 + \cos θ) (1 - \cos θ)} = \frac{2 (1 - \sin^{2} θ)}{2 (1 - \cos^{2} θ)} = \frac{\cos^{2} θ}{\sin^{2} θ} = \cot^{2} θ = \frac{225}{64}$

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Question 20:

In the given ∆ABC, it is given that ∠B = 90°, AB = 24 cm and BC = 7 cm. Then, cos A = ?
Figure

(a) $\frac{7}{24}$
(b) $\frac{7}{25}$
(c) $\frac{24}{25}$
(d) None of thse

Answer:

(c) $\frac{24}{25}$

Here, AB = 24 cm and BC = 7 cm
Now, using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC² = (24)² + (7)²
⇒ AC² = 625
⇒ AC = 25 cm
∴ cos A = $\frac{A B}{A C} = \frac{24}{25}$

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