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Trigonometry - Solved Examples

Exercise 8.4

Exercise 8.2

Trigonometry Concepts

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°.


$sin67° + cos75°$ = $sin(90° - 23°) + cos(90° - 75°)$ = $cos 23° + sin 15°$