Question

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

$\ln \left|\tan x\right|+\ln \left|\cot x\right|=0$

Short answer

$\ln \left|\tan x\right|+\ln \left|\cot x\right|=0$

Step-by-step solution

Step 1. Given information.

The Given equation, $\ln \left|\tan x\right|+\ln \left|\cot x\right|=0$.

Step 2.Convert left hand side in term of and and also apply logarithmic Given equation is: ln|tanx|+ln|cotx|=0

Simplifying L.H.S:

$\ln \left|\tan x\right|+\ln \left|\cot x\right|$

Putting this values in above equation we get,

$\begin{array}{l}\ln \left|\right(\tan x\left)\right|+\ln \left|\right(\cot x\left)\right|\\ =\ln \left|\left(\frac{\sin x}{\cos x}\right)\right|+\ln \left|\left(\frac{\cos x}{\sin x}\right)\right|\\ =0\end{array}$

Now using logarithmic identity in above step,

$\ln \left(\frac{x}{y}\right)=\ln \left(x\right)-\ln \left(y\right)$

$\ln \left|\right(\tan x\left)\right|+\ln \left|\right(\cot x\left)\right|=\ln \left|\right(\sin x)-\ln |\left(\cos x\right)|+\ln |\left(\cos x\right)|-\ln |\left(\sin x\right)\mid$

As terms with opposite signs cancel each other, so we get: L.H.S$=0$