Question

Proving Identities Involving Other Functions

These identities involve trigonometric functions as well as other functions that we have studied

Short answer

The expression $\ln \left|\tan x\sin x\right|=2\ln \left|\sin x\right|+\ln \left|\sec x\right|$is an identity.

Step-by-step solution

Step 1. Given information

An expression$\ln \left|\tan x\sin x\right|=2\ln \left|\sin x\right|+\ln \left|\sec x\right|$

Step 2. Concept used

A phrase is written in the area in issue. Divide it into LHS and RHS portions. Separately simplify them. It's an identity if both outcomes are the same.

Step 3. Calculation

Now, simplify LHS of $\ln \left|\mathrm{tanxsinx}\right|=2\ln \left|\mathrm{sinx}\right|+\ln \left|\mathrm{secx}\right|$

$\begin{array}{l}\ln \left|\tan x\sin x\right|=\ln \left|\frac{\sin x}{\cos x}\left(\sin x\right)\right|\\ =\ln \left|\frac{{\left(\sin x\right)}^2}{\cos x}\right|\\ =\ln \left|{\left(\sin x\right)}^2\times \sec x\right|\\ \because \ln \left(x^y\right)=y\ln \left(x\right)\&\ln (x\times y)=\ln \left(x\right)+\ln \left(y\right)\\ =2\ln \left|\sin x\right|+\left|\sec x\right|\end{array}$

RHS of the equation is$2\ln \left|\sin x\right|+\ln \left|\sec x\right|$

Both RHS and LHS are equal. Hence it is an identity