Introduction to Trigonometry ⇒⇒ Exercise 8.2

Question 1 (i)

Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°

Solution :

Given

sin 60° cos 30° + sin 30° cos 60°

As we know,

sin 30° = $ 1 \over 2 $, sin 60° = $ \sqrt{3} \over 2 $, cos 30° = $ \sqrt{3} \over 2 $, cos 60° = $ 1 \over 2 $.

By substituting trigonometric ratios

$$({\sqrt{3} \over 2} × {\sqrt{3} \over 2}) + ({ 1\over 2} × {1 \over 2})$$

$$⇒ {{3 \over 4} + {1 \over 4}}$$

$$⇒ {{3 + 1 }\over 4} $$

$$⇒ 1 $$

Therefore, sin 60° cos 30° + sin 30° cos 60° = 1.

Question 1 (ii)

Evaluate the following:
(ii)$$ 2 tan^2 45° + cos^2 30° – sin^2 60° $$

Solution :

Given

$$ 2 tan^2 45° + cos^2 30° – sin^2 60° $$

As we know,

sin 60° = $ \sqrt{3} \over 2 $, cos 30° = $ \sqrt{3} \over 2 $, tan 45° = 1.

By substituting trigonometric ratios

$$ 2 tan^2 45° + cos^2 30° – sin^2 60° $$

$$⇒ { 2 × 1 + ({3 \over 2})^2 - ({3 \over 2})^2}$$

$$⇒ 2 $$

Therefore, $ 2 tan^2 45° + cos^2 30° – sin^2 60° $ = 2.

Question 1 (iii)

Evaluate the following:
(iii) $$ cos 45° \over {sec 30° + cosec 30°}$$

Solution :

As we know,

sec 30° = $ 2 \over \sqrt{3} $, cosec 30° = 2 , cos 45° = $ 1 \over \sqrt 2 $

By substituting trigonometric ratios

$$ cos 45° \over {sec 30° + cosec 30°}$$

$$⇒ {{1 \over \sqrt 2} \over {{2 \over \sqrt{3}} + 2}}$$

$$⇒ {{1 \over \sqrt 2} \over {{2 + {2\sqrt{3} }\over \sqrt{3}}}}$$

$$⇒ {{1 \over \sqrt 2} × {{ \sqrt{3} \over 2 + {2\sqrt{3} }}}}$$

$$⇒ {\sqrt{3} \over {2(\sqrt 2) + {2\sqrt{6} }}}$$

Rationalising the denominator

$$ ⇒ {{\sqrt{3} \over {2\sqrt 2 + {2\sqrt{6} }}} × {{2\sqrt 2 - {2\sqrt{6} }} \over {2\sqrt 2 - {2\sqrt{6} }}}}$$

$$⇒ {{\sqrt{3}× ({{2 \sqrt 2 - {2\sqrt{6} }}})} \over {{(2\sqrt 2)^2 - {(2\sqrt{6})^2 }}}}$$

$$⇒ {{2 \sqrt 6 - {2\sqrt{18} }} \over {8 - 24}}$$

$$⇒ {{2 \sqrt 6 - {2\sqrt{18} }} \over {- 16}}$$

By taking –2 as common from numerator

$$ ⇒ { -2({ -\sqrt 6 + \sqrt{18}}) \over {- 2 × 8}}$$

$$⇒ {{\sqrt{(9 × 2)} -\sqrt 6 } \over { 8}}$$

$$ ⇒ {{3 \sqrt{ 2} -\sqrt 6 } \over { 8}}$$

Therefore, $ cos 45° \over {sec 30° + cosec 30°}$ = $ {3\sqrt{2} -\sqrt 6 } \over { 8}$.

Question 1 (iv)

Evaluate the following:
(iv) $$ {sin 30° + tan 45° − cosec60° } \over {sec 30° + cos 60° + cot 45°}$$

Solution :

As we know,

sin 30° = $ 1 \over 2 $, cosec 60° = $ 2 \over \sqrt{3} $ , tan 45° = $ 1 $

By substituting trigonometric ratios

$$ {sin 30° + tan 45° − cosec60° } \over {sec 30° + cos 60° + cot 45°}$$

$$⇒ {{{1 \over 2 }+ 1 − {2 \over \sqrt{3} }} \over {{2 \over \sqrt{3}} + {1 \over 2} + 1}}$$

$$⇒ {{{3 \over 2 } − {2 \over \sqrt{3} }} \over {{2 \over \sqrt{3}} + {3 \over 2}} }$$

$$⇒ {{{3\sqrt{3}- 4 } \over {2\sqrt{3}} } \over { 4 + {3\sqrt{3}} \over {2\sqrt{3}} }}$$

$$⇒ {{3\sqrt{3}- 4 } \over { 4 + {3\sqrt{3}}}}$$

Rationalising the denominator

$$ ⇒ {{ {3\sqrt{3}- 4 } × {( {3\sqrt{3}} - 4)} }\over {{( {3\sqrt{3}} + 4)} × {( {3\sqrt{3}} - 4)}}}$$

$$⇒ {{({3\sqrt{3}- 4 })^2}\over { ({3\sqrt{3}})^2 - 4^2 }}$$

$$⇒ {{27 - 24\sqrt{3} + 16 }\over { 27 - 16 }}$$

$$⇒ {{43 - 24\sqrt{3} }\over { 11 }}$$

Therefore, Answer : $$ {43 - 24\sqrt{3} }\over { 11 }$$.

Question 1 (v)

Evaluate the following:
(v) $$ {5 cos^2 60° + 4sec^2 30°- tan^2 45° } \over {sin^2 30° + cos^2 30°}$$

Solution :

As we know,

sin 30° = $ 1 \over 2 $, cos 30° = $ \sqrt{3} \over 2 $ , tan 45° = $ 1 $.

sec 30° = $ 2 \over \sqrt{3} $, cos 60° = $ 1 \over 2 $.

By substituting trigonometric ratios

$$ {5 cos^2 60° + 4sec^2 30°- tan^2 45° } \over {sin^2 30° + + cos^2 30°}$$

$$ {5({1 \over 2 })^2 + {4 ({2 \over \sqrt{3}})}^2 - {1}^2} \over {({1 \over 2 })^2 + ({\sqrt{3} \over 2})^2 }$$

$$⇒ {{5 × {1 \over 4} + {4 × {4 \over 3}} - 1} \over {{1 \over 4 } + {3 \over 4} }}$$

$$⇒ {{ {5 \over 4} + { {16 \over 3}} - 1} \over {{1 +3 } \over 4 }}$$

$$⇒ { {15 + 64 - 12 }\over 12 } × {4 \over 4 }$$

$$⇒ { 67 \over 12 } $$

Therefore, Answer : $$ { 67 \over 12 } $$.

Question 2 (i)

Choose the correct option and justify your choice:
(i) $$ {{2tan 30° } \over { 1 + tan^2 30°} } = $$
(A) sin 60° (B) cos 60°
(C) tan 60° (D) sin 30°

Solution :

As we know,

tan 30° = $ {1 \over \sqrt{3}} $.

By substituting trigonometric ratios

$$ {{2tan 30° } \over { 1 + tan^2 30°} } $$

$$⇒ {{2 × {1 \over \sqrt{3}} } \over { 1 + ({1 \over \sqrt{3}})^2 } } $$

$$⇒ {{2 \over \sqrt{3}} \over { 1 + {1 \over 3 } }} $$

$$⇒ {{2 \over \sqrt{3}} \over { 4 \over 3 } } $$

$$⇒ {2 \over \sqrt{3}} × {3 \over 4 }$$

$$⇒ { \sqrt{3} \over 2 } $$

Only sin 60° has the value equal to $ { \sqrt{3} \over 2 } $

Therefore, option (A) sin 60° is the correct option.

Question 2 (ii)

Choose the correct option and justify your choice:
(ii) $$ {{ 1 - tan^2 45°} \over { 1 + tan^2 45°} } = $$
(A) tan 90° (B) 1
(C) sin 45° (D) 0

Solution :

As we know,

tan 45° = $ 1 $.

By substituting trigonometric ratios

$$ {{ 1 - tan^2 45°} \over { 1 + tan^2 45°} } $$

$$ ⇒ {{ 1 - 1} \over { 1 + 1} }$$

$$⇒ {0 \over { 2 } } $$

$$⇒ 0 $$

Therefore, option (D) is the correct option.

Question 2 (iii)

Choose the correct option and justify your choice:
(iii) sin 2 A = 2 sin A is true when A =
(A) 0° (B) 30°
(C) 45° (D) 60°

Solution :

Given,

sin 2A = 2 sin A

Substituting A = 0° in L.H.S. and R.H.S. of above equation :

$$ {sin(0) = 2sin 0 } $$

$$ ⇒ {0 = 0 }$$

$$⇒ {L.H.S = R.H.S } $$

Therefore, option (A) is the correct option.

Question 2 (iv)

Choose the correct option and justify your choice:
(iv) $$ {{ 2 tan 30°} \over { 1 - tan^2 30°} } = $$
(A) cos 60° (B) sin 60°
(C) tan 60° (D) sin 30°

Solution :

As we know,

tan 30° = $ {1 \over \sqrt{3}} $.

By substituting trigonometric ratios

$$ {{2tan 30° } \over { 1 - tan^2 30°} } $$

$$⇒ {{{2 × {1 \over \sqrt{3}} } \over { 1 - ({1 \over \sqrt{3}})^2 } }} $$

$$⇒ {{2 \over \sqrt{3}} \over { 1 - {1 \over 3 } } } $$

$$⇒ {{2 \over \sqrt{3}} \over { 2 \over 3 } } $$

$$⇒ {{2 \over \sqrt{3}} × {3 \over 2 }}$$

$$⇒ { \sqrt{3} } $$

Only tan 60° has the value equal to $ \sqrt{3} $

Hence, option (C) is correct.

Question 3

If tan (A+B) = $ \sqrt3 $ and tan (A – B) = $ {1 \over \sqrt{3}}$
0° < A + B ≤ 90°; A > B,
find A and B.

Solution :

Given, tan (A + B) = √3

As we know,

tan 60° = √3

$$⇒ A + B = 60° .....(i)$$

Given, tan (A - B) = $ {1 \over \sqrt{3}} $

As we know,

tan 30° = $ {1 \over \sqrt{3}} $

$$⇒ A - B = 30° .....(ii)$$

On adding both equations (i) and (ii), we obtain:

$$⇒ A + B + A – B = 60° + 30°$$

$$⇒ 2A = 90° $$

$$⇒ A = 45° $$

Put A = 45° in equation (1)

$$⇒ A + B = 60°$$

$$⇒ 45° + B = 60° $$

$$⇒ B = 60° - 45° $$

$$⇒ B = 15° $$

Thus, A = 45° and B = 15°

Question 4 (i)

State whether the following are true or false. Justify your answer.
(i) sin (A+B) = sin A + sin B

Solution :

For the purpose of verification, Let

A = 45° and B = 45°

LHS: sin (45° + 45°)

$$⇒ sin 90° = 1 $$

RHS : sin 45° + sin 45°

$$ {1 \over \sqrt{2}} + {1 \over \sqrt{2}} $$

$$⇒ {2 \over \sqrt{2}} $$

⇒ LHS ≠ RHS.

Since, sin (A + B) ≠ sin A + sin B. Given statement is False .

Question 4 (ii)

State whether the following are true or false. Justify your answer.
(ii) The value of sin θ increases as θ increases

Solution :

For the purpose of verification, Let θ be 0°, 30° and 45°, 90°

$$ sin 0° = 0 $$

$$ sin 30° = { 1 \over 2 }$$

$$ sin 45° = {1 \over {{\sqrt 2}} } = 0.707 $$

$$ sin 60° = {\sqrt{3} \over 2 }= 0.866 $$

$$ sin 90° = 1$$

Hence, the statement the value of sin θ increases as θ increases is true.

Question 4 (iii)

State whether the following are true or false. Justify your answer.
(iii) The value of cos θ increases as θ increases

Solution :

For the purpose of verification, Let θ be 0°, 30° and 45°, 90°

$$ cos 0° = 1 $$

$$ cos 30° = {\sqrt{3} \over 2 }= 0.866 $$

$$ cos 45° = {1 \over {{\sqrt 2}} } = 0.707 $$

$$ cos 60° = { 1 \over 2 }$$

$$ cos 90° = 0 $$

Hence, the statement the value of cos θ increases as θ increases is False.

Question 4 (iv)

State whether the following are true or false. Justify your answer.
(iv) sin θ = cos θ for all values of θ.

Solution :

For the purpose of verification, Let θ = 0°

sin 0° = 0 and cos 0° = 1

From the values of standard angles tables it is wrong. It is true only for 45°.

Hence, the statement sin θ = cos θ for all values of θ is False.

Question 4 (v)

State whether the following are true or false. Justify your answer.
(v) cot A is not defined for A =0°.

Solution :

As we know

cot A = cos A / sin A

cot 0° = cos 0° / sin 0°

cot 0° = 1 / 0 = $ \infty = Undefined $

Hence, the statement cot A is not defined for A = 0° is true.