What is the value of \frac{\tan^2x-\sin^2x}{\sec^2x} ?

SSC CHSL Model Paper 15

Question : 70 of 100

Marks: +1, -0

What is the value of

{\tan^2x-\sin^2x}/{\sec^2x}

Solution:

{\tan^2x-\sin^2x}/{\sec^2x}

={ {\sin^2x}/{\cos^2x}-\sin^2x}/{\sec^2x}

={\sin^2x({1-\cos^2x}/{\cos^2x})}/{\sec^2x}

={\sin^2x \sin^2x}/{\cos^2x \sec^2x}=\sin^4x

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