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Introduction to Trigonometry : Exercise - 8.2 (Mathematics NCERT Class 10th)

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Q.1     Evaluate :
(i) sin 60 ° cos 30° + sin 30° cos 60°
(ii) $2{\tan ^2}45^\circ + {\cos ^2}30^\circ - {\sin ^2}60^\circ$
(iii) ${{\cos 45^\circ } \over {\sec 30^\circ + \cos ec30^\circ }}$
(iv) ${{\sin 30^\circ + \tan 45^\circ - \cos ec60^\circ } \over {\sec 30^\circ + \cos 60^\circ + \cot 45^\circ }}$
(v) ${{5{{\cos }^2}60^\circ + 4{{\sec }^2}30^\circ - {{\tan }^2}45^\circ } \over {{{\sin }^2}30^\circ + {{\cos }^2}30^\circ }}$
Sol.     (i) sin 60° cos 30° + sin 30° cos 60°
$= {{\sqrt 3 } \over 2} \times {{\sqrt 3 } \over 2} + {1 \over 2} \times {1 \over 2}$

$= {3 \over 4} + {1 \over 4} = {{3 + 1} \over 4} = {4 \over 4} = 1$

(ii) $2{\tan ^2}45^\circ + {\cos ^2}30^\circ - {\sin ^2}60^\circ$
$= 2{\left( 1 \right)^2} + {\left( {{{\sqrt 3 } \over 2}} \right)^2} - {\left( {{{\sqrt 3 } \over 2}} \right)^2}$
$= 2 + {3 \over 4} - {3 \over 4} = 2$

(iii) ${{\cos 45^\circ } \over {\sin 30^\circ + \cos ec30^\circ }}$
$= {{{1 \over {\sqrt 2 }}} \over {{2 \over {\sqrt 3 }} + 2}} = {{{1 \over {\sqrt 2 }}} \over {2 + {{2\sqrt 3 } \over {\sqrt 3 }}}} = {1 \over {\sqrt 2 }} \times {{\sqrt 3 } \over {2 + 2\sqrt 3 }}$
= ${{\sqrt 3 } \over {\sqrt 2 \times 2\left( {\sqrt 3 + 1} \right)}} \times {{\sqrt 3 - 1} \over {\sqrt 3 - 1}}$
= ${{\sqrt 3 \left( {\sqrt 3 - 1} \right)} \over {\sqrt 2 \times 2\left( {3 - 1} \right)}} = {{\sqrt 2 \times \sqrt 3 \left( {\sqrt 3 - 1} \right)} \over {\sqrt 2 \times \sqrt 2 \times 2 \times 2}}$
= ${{3\sqrt 2 - \sqrt 6 } \over 8}$

(iv) ${{\sin 30^\circ + \tan 45^\circ - \cos ec60^\circ } \over {\sin 30^\circ + \cos 45^\circ + \cot 45^\circ }}$
${{{1 \over 2} + 1 - {2 \over {\sqrt 3 }}} \over {{2 \over {\sqrt 3 }} + {1 \over 2}}} = {{{{1 + 2} \over 2} - {2 \over {\sqrt 3 }}} \over {{2 \over {\sqrt 3 }} + {{1 + 2} \over 2}}}{\mkern 1mu}$
$= {{{3 \over 2} - {2 \over {\sqrt 3 }}} \over {{2 \over {\sqrt 3 }} + {3 \over 2}}} = {{{{3\sqrt 3 - 4} \over {2\sqrt 3 }}} \over {{{4 + 3\sqrt 3 } \over {2\sqrt 3 }}}}$
$= {{3\sqrt 3 - 4} \over {4 + 3\sqrt 3 }} = {{\left( {3\sqrt 3 - 4} \right)\left( {3\sqrt 3 - 4} \right)} \over {\left( {4 + 3\sqrt 3 } \right)\left( {3\sqrt 3 - 4} \right)}}$
${{27 + 16 - 12\sqrt 3 - 12\sqrt 3 } \over {27 - 16}}$
${{43 - 24\sqrt 3 } \over {11}}$

(v) ${{5{{\cos }^2}60^\circ + 4{{\sec }^2}30^\circ - {{\tan }^2}45^\circ } \over {{{\sin }^2}30^\circ + {{\cos }^2}30^\circ }}$
Sol.       ${{5{{\left( {{1 \over 2}} \right)}^2} + 4{{\left( {{2 \over {\sqrt 3 }}} \right)}^2} - {{\left( 1 \right)}^2}} \over {{{\left( {{1 \over 2}} \right)}^2} + {{\left( {{{\sqrt 3 } \over 2}} \right)}^2}}}$
${{5 \times {1 \over 4} + 4 \times {4 \over 3} - 1} \over {{1 \over 4} + {3 \over 4}}} = {5 \over 4} + {{16} \over 3}-1 = {{{1 \over {12}}\left( {15 + 64 - 12} \right)} \over {{{1 + 3} \over 4}}}$
${{{1 \over {12}} \times 67} \over {{4 \over 4}}} = {{67} \over {12}}$

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

(ii) ${{1 - {{\tan }^2}45^\circ } \over {1 + {{\tan }^2}45^\circ }} =$
(A) tan 90°                   (B) 1
(C) sin 45 °                   (D) 0

(iii) sin 2A = 2 sinA is true when A =
(A) 0°                           (B) 30°
(C) 45°                         (D) 60°

(iv) ${{2{{\tan }^2}30^\circ } \over {1 - {{\tan }^2}30^\circ }} =$
(A) cos 60°                   (B) sin 60°
(C) tan 60 °                  (D) none of these
Sol.        (i) (A)
Because ${{2\tan 30^\circ } \over {1 + {{\tan }^2}30^\circ }} = {{2 \times {1 \over {\sqrt 3 }}} \over {1 + {{\left( {{1 \over {\sqrt 3 }}} \right)}^2}}} = {{{2 \over {\sqrt 3 }}} \over {1 + {1 \over 3}}}$
$= {2 \over {\sqrt 3 }} \times {3 \over {3 + 1}} = {2 \over {\sqrt 3 }} \times {3 \over 4}$
$= {{\sqrt 3 } \over 2} = \sin 60^\circ$

(ii) (D)
Because ${{1 - {{\tan }^2}45^\circ } \over {1 + {{\tan }^2}45^\circ }} = {{1 - 1} \over {1 + 1}} = {0 \over 2} = 0$

(iii) (A)
Because when A = 0, sin 2 A = sin 0 = 0
and, 2 sinA = 2 sin 0 = 2 × 0 = 0
$\Rightarrow$ sin 2A = 2sinA, when A = 0

(iv) (C)
Because ${{2\tan 30^\circ } \over {1 - {{\tan }^2}30^\circ }} = {{2 \times {1 \over {\sqrt 3 }}} \over {1 - {{\left( {{1 \over {\sqrt 3 }}} \right)}^2}}} = {{{2 \over {\sqrt 3 }}} \over {1 - {1 \over 3}}}$
$= {2 \over {\sqrt 3 }} \times {3 \over {3 - 1}} = {2 \over {\sqrt 3 }} \times {3 \over 2}$
$= \sqrt 3 = \tan 60^\circ$

Q.3     If tan (A + B) = $\sqrt 3$ and tan (A – B) = ${1 \over {\sqrt 3 }};0^\circ < A + B \le 90^\circ ;A > B,$ find A and B.
Sol.       tan (A + B) = ${\sqrt 3 }$
$\Rightarrow \tan \left( {A + B} \right) = \tan 60^\circ$
$\Rightarrow A + B = 60^\circ$ ...(1)
$\tan \left( {A - B} \right) = {1 \over {\sqrt 3 }}$
$\Rightarrow \tan \left( {A - B} \right) = \tan 30^\circ$
$\Rightarrow$ A – B = 30 °...(2)
Solving (1) and (2), we get
A = 45° and B = 15°
Hence, A = 45° and B = 15°

Q.4     State whether the following are true of false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin $\theta$ increases as $\theta$ increases.
(iii) The value of cas$\theta$ increases as $\theta$ increases.
(iv) sin$\theta$ = cos$\theta$ for all values of $\theta$.
(v) cot A is not defined for A = 0°.
Sol.       (i) False Because
When A = 60° and B = 30°. Then,
sin (A + B) = sin (60° + 30°) = sin 90° = 1
and, sin A + sin B = sin 60° + sin 30°
$= {{\sqrt 3 } \over 2} + {1 \over 2} = {{\sqrt 3 + 1} \over 2}$
$\Rightarrow$ sin (A + B) $\ne$ sin A + sin B
(ii) True Because
Clearly from the table below : We find that the value of sin$\theta$ increases as $\theta$ increases.

(iii) False Because
Clearly from the table below : We find that the value of cos$\theta$ decreases as $\theta$ increases.

(iv)  False as it is only true for $\theta$ = 45°.
$\left( {\sin 45^\circ = {1 \over {\sqrt 2 }} = \cos 45^\circ } \right)$

(v) True Because tan 0° = 0 and $\cot 0^\circ = {1 \over {\tan 0^\circ }} = {1 \over 0},i.e.,$ undefined.

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