NCERT Exercise 8.3
1
Question 1:
Evaluate :
(i) $\dfrac{\text{sin 18°}}{\text{cos 72°}}$
(ii) $\dfrac{\text{tan 26°}}{\text{cot 64°}}$
(iii) $\text{cos 48°} - \text{sin 42°}$
(iv) $\text{cosec 31°} - \text{sec 59°}$
Solution
(i) $\dfrac{\text{sin 18°}}{\text{cos 72°}}$
$= \dfrac{\text{sin (90° - 72°)}}{\text{cos 72°}}$
$= \dfrac{\text{cos 72°}}{\text{cos 72°}}$ = 1
(ii) $\dfrac{\text{tan 26°}}{\text{cot 64°}}$
$= \dfrac{\text{(tan 90° - 64°)}}{\text{cot 64°}}$
$\dfrac{\text{cot 64°}}{\text{cot 64°}}$ = 1
(iii) $\text{cos 48°} - \text{sin 42°}$
$= \text{cos (90° - 42°)} - \text{sin 42°}$
$= \text{sin 42°} - \text{sin 42°}$
= 0
(iv) $\text{cosec 31°} - \text{sec 59°}$
$= \text{cosec (90° - 59°)} - \text{sec 59°}$
$= \text{sec 59°} - \text{sec 59°}$
= 0
2
Show that:
(i) $tan48°tan23°tan42°tan67°$ = 1
(ii) $cos38°cos52° - sin38°sin52°$ = 0
Solution
(i) $tan48°tan23°tan42°tan67°$
$= tan(90° - 42°)tan(90° - 67°)tan42°tan67°$
$= cot48°cot23°tan42°tan67°$ = 1
(ii) $cos38°cos52° - sin38°sin52°$ = 0
$= cos(90° - 52°)cos(90° - 38°) - sin38°sin52°$
$= sin52°sin38° - sin38°sin52°$ = 0
3
If tan2A = cot(A - 18°), where 2A is an acute angle, find the value of A.
Solution
tan 2A = cot (A − 18°)
∴ cot (90° − 2A) = cot (A − 18°)
∴ 90° − 2A = A − 18°
⇒ 108° = 3A ⇒ A = 36°
4
If tan A = cot B, prove that A + B = 90°
tan A = cot B
∴ tan A = tan (90° − B)
⇒ A = 90° − B ⇒ A + B = 90°
5
If sec4A=cosec(A - 20°), where 4A is an acute angle, find the value of A.
sec 4A = cosec (A − 20°)
⇒ cosec (90° − 4A) = cosec (A − 20°)
∴ 90° − 4A = A− 20°
⇒ 110° = 5A
⇒ A = 22°
6
If A, B and C are interior angles of a triangle ABC, then show that
$sin\left(\dfrac{B+C}{2}\right)$ = $cos\dfrac{A}{2}$
In a △ ABC, $ A + B + C = 180°$
⇒ $B + C = 180° - A$
⇒ $\left(\dfrac{B + C}{2}\right)$ = $\left(\dfrac{180° - A}{2}\right)$
⇒ $sin\left(\dfrac{B + C}{2}\right)$ = $sin\left(\dfrac{180° - A}{2}\right)$
⇒ $sin\left(\dfrac{B + C}{2}\right)$ = $sin\left(90° - \dfrac{A}{2}\right)$
⇒ $sin\left(\dfrac{B + C}{2}\right)$ = $cos\left(\dfrac{A}{2}\right)$
7
Express sin67° + cos75° in terms of trigonometric ratios of angles between 0° and 45°.
$sin67° + cos75°$ = $sin(90° - 23°) + cos(90° - 75°)$ = $cos 23° + sin 15°$