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Question Number 57345 by behi83417@gmail.com last updated on 02/Apr/19

1) if : sinx + tgx = 1, then : \sin^{4} x + {tg}^{4} x = ?

2) if : sinx + tgx = 2, then : \sin 4 x + tg 4 x = ?

3 . if : sinx + tgx = 3, then : \frac{\sin 4 x}{\sin^{4} x} + \frac{tg 4 x}{{tg}^{4} x} = ?

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Apr/19

s i n x + \frac{s i n x}{c o s x} = a

\frac{2 t}{1 + t^{2}} + \frac{2 t}{1 + t^{2}} \times \frac{1 + t^{2}}{1 - t^{2}} = a

\frac{2 t}{1 + t^{2}} + \frac{2 t}{1 - t^{2}} = a

\frac{2 t - 2 t^{3} + 2 t + 2 t^{3}}{1 - t^{4}} = a

\frac{4 t}{1 - t^{4}} = a

a - {a t}^{4} - 4 t = 0

{a t}^{4} + 4 t - a = 0

t = t a n (\frac{x}{2})

w h e n a = 1 t = 0.249

w h e n a = 2 t = 0.475

w h e n a = 3 t = 0.631

r e s t o f t h e p r o b l e m c a n b e s o l v e d b y

p u t t i n g \frac{2 t}{1 + t^{2}} f o r s i n x a n d \frac{2 t}{1 - t^{2}} f o r t a n x

t h e n p u t v a l u e o f t