Find the value of: (i) sin 75° (ii) tan 15°

Solution:

(i) sin 75°

sin 75° = sin (45º + 30º)

= sin 45° cos 30º + cos 45° sin 30°     [ ∵ sin (x+y) = sin x cos y + cos x sin y]

= (1/√2) × (√3/2) + (1/√2) × (1/2) (by trigonometric table)

= √3/(2√2) + 1/(2√2)

= (√3+1)/(2√2)

(ii) tan 15°

tan 15° = tan (45º - 30º)

= (tan 45º - tan 30º) / (1 + tan 45° tan 30°)   

 [∵ tan (x-y) = (tan x - tan y) / (1 + tan x tan y)]

= [1-(1/√3)] / [1+ 1 × (1/√3)] (by trigonometric table)

= [(√3-1)/√3] / [√3+1/√3]

= (√3-1)/(√3+1)

= [(√3-1)/(√3+1)] × [(√3-1)/(√3-1)]     [By rationalizing the denominator]

= (√3-1)² / (3-1)

= (3+1-2√3) / 2

= 2(2-√3)/2

= 2-√3

NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3 Question 5

Find the value of: (i) sin 75° (ii) tan 15°

Summary:

The value of (i) sin 75° = (√3+1)/(2√2) and (ii) tan 15° = 2-√3