We have, sin #10;^4xcos^2x #10; #10; =sin #10;^2xsin^2xcos^2x #10; #10; =sin #10;^2xtan^2x #10; #10;Hence, this is the answer. ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Trigonometric Identities

Solve: sin4x...

Question

Solve:
$\frac{{sin}^{4} x}{{cos}^{2} x}$

Open in App

Solution

We have,
$\frac{{sin}^{4} x}{{cos}^{2} x}$

$= \frac{{sin}^{2} x {sin}^{2} x}{{cos}^{2} x}$

$= {sin}^{2} x {tan}^{2} x$

Hence, this is the answer.

Suggest Corrections

Similar questions

\cos x + \cos 2 x + \cos 3 x = 0

\cos x + \cos 3 x - \cos 2 x = 0

\sin x + \sin 5 x = \sin 3 x

\cos x \cos 2 x \cos 3 x = \frac{1}{4}

\cos x + \sin x = \cos 2 x + \sin 2 x

\sin x + \sin 2 x + \sin 3 = 0

\sin x + \sin 2 x + \sin 3 x + \sin 4 x = 0

\sin 3 x - \sin x = 4 \cos^{2} x - 2

\sin 2 x - \sin 4 x + \sin 6 x = 0

s i n (4 x + \frac{π}{4}) + c o s (4 x + \frac{5 π}{4}) = \sqrt{2}

\frac{c o s 2 x}{c o s 2 - s i n 2} > 2 - s i n 4 x .

\int_{0}^{π / 4} \frac{sin x cos x d x}{{cos}^{2} x + {sin}^{4} x}

\int \frac{1}{{sin}^{4} x + {sin}^{2} x {cos}^{2} x + {cos}^{4} x} d x

\int \frac{cos 4 x - cos 2 x}{sin 4 x - sin 2 x} d x

Join BYJU'S Learning Program