Question

Proving Identities Involving Other Functions These identities involve trigonometric functions as well as other functions that we have studied.

$e^{{\sin }^{2}x}{e}^{{\tan }^{2}x}=e^{{\sec }^{2}x}{e}^{-{\cos }^{2}x}$

Short answer

$e^{{\sin }^{2}x}{e}^{{\tan }^{2}x}=e^{{\sec }^{2}x}{e}^{-{\cos }^{2}x}$

Step-by-step solution

Step 1. Given information.

The given equaution $e^{{\sin }^{2}x}{e}^{{\tan }^{2}x}=e^{{\sec }^{2}x}{e}^{-{\cos }^{2}x}$,

Step 2. Convert left hand side in term of and and also apply logarithmic identity and proceed

Given equation is:${\text{e}}^{{\sin }^{2}x}\cdot e^{{\tan }^{2}x}=e^{{\sec }^{2}x}\cdot e^{-{\cos }^{2}x}$

Consider R.H.S.As

$\begin{array}{l}{\sec }^{2}x-{\tan }^{2}x=1\\ \therefore {\sec }^{2}x=1+{\tan }^{2}x\end{array}$

Putting this values in above equation,

$e^{{\sec }^{2}x}\cdot e^{-{\cos }^{2}x}=e^{1+{\tan }^{2}x}\cdot e^{-{\cos }^{2}x}$

Now using,

$\begin{array}{l}{e}^{{\sec }^{2}x}\cdot e^{-{\cos }^{2}x}=e^{1+{\tan }^{2}x-{\cos }^{2}x}\\ \text{As,}1-{\cos }^{2}x={\sin }^{2}x\\ e^{{\sec }^{2}x}\cdot e^{-{\cos }^{2}x}=e^{{\tan }^{2}x+{\sin }^{2}x}\\ e^{{\sec }^{2}x}\cdot e^{-{\cos }^{2}x}=e^{{\tan }^{2}x}\cdot e^{{\sin }^{2}x}\\ {\text{(Ase}}^x{\text{e}}^y=e^{x+y}\text{)}\end{array}$