Introduction To Trigonometry Exercise 8.3

Q.1     Evaluate :
(i) ${{\sin 18^\circ } \over {\cos 72^\circ }}$
(ii) ${{\tan 26^\circ } \over {\cot 64^\circ }}$
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°
Sol.       (i) ${{\sin 18^\circ } \over {\cos 72^\circ }} = {{\sin \left( {90^\circ - 72} \right)} \over {\cos 72^\circ }} = {{\cos 72^\circ } \over {\cos 72^\circ }} = 1[Since,\sin 90\,\,-\theta = \cos \theta ]$

(ii) ${{\tan 26^\circ } \over {\cot 64^\circ }} = {{\tan \left( {90^\circ - 64^\circ } \right)} \over {\cot 64^\circ }} = {{\cot 64^\circ } \over {\cot 64^\circ }} = 1$ $[Since,\tan \left( {90 - \theta } \right) = \cot \theta ]$

(iii) cos48° – sin42° = cos (90° – 42°) – sin 42°
= sin 42° – sin 42° = 0 $[Since,\cos \left( {90 - \theta } \right) = \sin \theta ]$
(iv) cosec 31° – sec 59° = cosec (90° – 59°) – sec 59° = 0
$= \sec 59^\circ - \sec 59^\circ = 0$ $[Since,\cos ec\left( {60 - \theta } \right) = \sec \theta ]$

Q.2     Show that
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52 ° = 0
Sol.       (i) tan 48° tan 23° tan 42 ° tan 67°
= tan (90° – 42°) tan(90° – 67°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°  $[Since,\tan \left( {90 - \theta } \right) = \cot \theta ]$
$= {1 \over {\tan 42^\circ }}.{1 \over {\tan 67^\circ }}.\tan 42^\circ \tan 67^\circ = 1$

(ii) cos 38°cos 52°– sin 38°sin 52°
= cos(90° – 52°)cos(90°– 38°) – sin 38°sin 52°
= sin 52° sin 38° – sin 38°sin 52° = 0 $[Since,\cos \left( {90^\circ - \theta } \right) = \sin \theta ]$

Q.3     If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Sol.       We are given that
tan 2A = cot (A – 18°)...(1)
Since tan 2A = cot (90° – 2A), so we can write (1) as
cot(90°–2A) = cot (90° – 2A), so we can write (1) as
cot (90° – 2A) = cot (A – 18°) $[Since,\cot \left( {90^\circ - \theta } \right) = \tan \theta ]$
Since (90° – 2A) and (A – 18°) are both acute angle therefore,
90° – 2A = A – 18°
$\Rightarrow$ – 2A – A = – 18° – 90°
$\Rightarrow$ – 3A = – 108°
$\Rightarrow$ A = 36°

Q.4     If tan A = cot B, prove that A + B = 90°.
Sol.       We are given that
tan A = cot B...(1)
Since tan A = cot(90° – A), so we can write (1) as
cot (90° – A ) = cot B $[Since,\cot \left( {90^\circ - \theta } \right) = \tan \theta ]$
Since (90° – A) and B are both acute angles, therefore,
(90° – A) = B
$\Rightarrow$ A + B = 90°

Q.5     If sec 4 A = cosec (A – 20°), where 4 A is an acute angle, find the value of A.
Sol.      We are given that sec 4A = cosec (A – 20°). ....(1)
Since sec 4A = cosec (90°– 4A), so we can write (1) as
cosec (90° – 4A) = cosec (A – 20°) $[Since,\cos ec\left( {90^\circ - \theta } \right) = \sec \theta ]$
Since (90° – 4A) and (A – 20°) are both acute angles, therefore,
90° – 4A = A – 20°
$\Rightarrow$ – 4A – A = – 20° – 90°
$\Rightarrow$ – 5A = 110°
$\Rightarrow$ A = 22°

Q.6     If A, B and C are interior angles of a $\Delta ABC,$ then show that
$\sin \left( {{{B + C} \over 2}} \right) = \cos {A \over 2}$
Sol.       Since A, B and C are the interior angles of a $\Delta ABC,$ therefore,
A + B + C = 180°
$\Rightarrow$ ${{A + B + C} \over 2} = 90^\circ$
$\Rightarrow$ ${{B + C} \over 2} = 90^\circ - {A \over 2}$
$sin\left( {{{B + C} \over 2}} \right) = 90^\circ - {A \over 2}$
$\sin \left( {{{B + C} \over 2}} \right) = \cos {A \over 2}$ $[Since,\sin \left( {90^\circ - \theta } \right) = \cos \theta ]$

Q.7     Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Sol.       sin 67° + cos 75° = sin (90° – 23) + cos (90° – 15°)
= cos 23° + sin 15°

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