$\frac{s i n x - s i n 3 x + s i n 5 x - s i n 7 x}{c o s x - c o s 3 x - c o s 5 x + c o s 7 x} = c o t 2 x$

Open in App

Solution

We have,

L.H.S.

$\frac{sin x - sin 3 x + sin 5 x - sin 7 x}{cos x - cos 3 x - cos 5 x + cos 7 x}$

$\Rightarrow \frac{(sin x - sin 3 x) + (sin 5 x - sin 7 x)}{(cos x - cos 3 x) - (cos 5 x - cos 7 x)}$

$\Rightarrow \frac{2 cos \frac{(x + 3 x)}{2} sin \frac{(x - 3 x)}{2} + 2 cos \frac{(5 x + 7 x)}{2} sin \frac{(5 x - 7 x)}{2}}{2 sin \frac{(x + 3 x)}{2} sin \frac{(3 x - x)}{2} - 2 sin \frac{(5 x + 7 x)}{2} sin \frac{(7 x - 5 x)}{2}}$

$\Rightarrow \frac{2 cos 2 x (- sin x) + 2 cos 6 x (- sin x)}{2 sin 2 x sin x - 2 sin 6 x sin x}$

$\Rightarrow \frac{- 2 sin x (cos 2 x + cos 6 x)}{2 sin x (sin 2 x - sin 6 x)}$

$\Rightarrow \frac{- (cos 2 x + cos 6 x)}{(sin 2 x - sin 6 x)}$

$\Rightarrow \frac{- (2 cos \frac{2 x + 6 x}{2} cos \frac{2 x - 6 x}{2})}{2 cos \frac{2 x + 6 x}{2} sin \frac{2 x - 6 x}{2}} ∴ cos (- θ) = cos θ$

$\Rightarrow \frac{- cos 2 x}{- sin 2 x}$

$\Rightarrow cot 2 x$

Suggest Corrections

Similar questions

\frac{sin x - sin 3 x + sin 5 x - sin 7 x}{cos x - cos 3 x - cos 5 x + cos 7 x} = cot 2 x

\frac{(sin 7 x + sin 5 x) + (sin 9 x + sin 3 x)}{(cos 7 x + cos 5 x) + (cos 9 x + cos 3 x)} = tan 6 x

\frac{sin x - sin 3 x + sin 5 x - sin 7 x}{cos x - cos 3 x - cos 5 x + cos 7 x} = cot 2 x

y = \frac{cos x + cos 2 x + cos 3 x + cos 4 x + cos 5 x + cos 6 x + cos 7 x}{sin x + sin 2 x + sin 3 x + sin 4 x + sin 5 x + sin 6 x + sin 7 x},

\frac{sin 5 x + sin 3 x}{cos 5 x + cos 3 x} =

Join BYJU'S Learning Program