Answer
$e^{\sin^2 x}e^{\tan^2 x}=e^{\sec^2 x} e^{-\cos^2 x}$
Work Step by Step
Need to verify $e^{\sin^2 x}e^{\tan^2 x}=e^{\sec^2 x} e^{-\cos^2 x}$
Consider left hand side : $e^{\sin^2 x}e^{\tan^2 x}=e^{\sin^2 x+\tan^2 x}$...(1)
Now, $\sin^2 x+\tan^2 x=\sin^2 x+\dfrac{\sin^2x}{\cos^2 x}$
or, $\dfrac{1-\cos^4 x}{\cos^2 x}=\dfrac{1}{\cos^2 x}-\cos^2 x$
Thus, equation (1) becomes:
$e^{\sin^2 x}e^{\tan^2 x}=e^{\sin^2 x+\tan^2 x}=\dfrac{1}{\cos^2 x}-\cos^2 x$
$e^{\sin^2 x}e^{\tan^2 x}=e^{\sec^2 x-\cos^2 x}=e^{\sec^2 x} e^{-\cos^2 x}$ (RHS)
Hence, the left-hand side and right hand side are equal.
