Find sin x/2, cos x/2, and tan x/2 in each of the following: tan x = - 4/3, x in quadrant II

Solution:

Since x lies in quadrant II i.e. π/2 < x < π

Therefore, π/4 < x/2 < π/2

Hence, sin x/2, cos x/2, and tan x/2 are positive as x/2 lies in quadrant I.

It is given that tan x = - 4/3

By one of the trigonometric identities,

sec²x = 1 + tan²x

= 1 + (-4/3)²

= 1 + 16/9

= 25/9

sec x = ± √(25/9)

1 / cos x = ± 5/3

cos x = ± 3/5

As x is in quadrant II, cos x is negative

cos x = - 3/5

cos 2(x/2) = -3/5

By double angle formulas, 

2cos²(x/2) - 1 = - 3/5

2cos²x/2 = 1 - 3/5

cos²x/2 = (2/5) × (1/2)

cos²x/2 = 1/5

cos x/2 = ± √(1/5)

Since cos x/2 lies in quadrant I and positive so, cos x/2 = 1/√5 (or) √5/5

Now, sin²x/2 = 1 - cos²x/2 [Because sin²A + cos ²A = 1]

= 1 - (1/√5)²

= 1 - 1/5

= 4/5

sin x/2 = ± √(4/5)

sin x/2 = ± 2/√5

Since sin x/2 lies in quadrant I and positive so, sin x/2 = 2/√5 (or) 2√5/5

Now, tan x/2 = [sin x/2] / [cos x/2]

= (2/√5) / (1/√5)

= 2

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NCERT Solutions Class 11 Maths Chapter 3 Exercise ME Question 8

Find sin x/2, cos x/2, and tan x/2 in each of the following: tan x = - 4/3, x in quadrant II

Summary:

When tan x = -4/3 and x is in quadrant II, sin x/2 = (2√5/5), cos x/2 = (√5/5), and tan x/2 = 2