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. ...

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Trigonometric Identities

Solve: sin4x...

Question

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

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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.

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