Prove:
$ \left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A $

Given:


$\frac{1+tan^2A}{1+cot^2A}=(\frac{1-tanA}{1-cotA})^2=tan^2A$

To do:

We have to prove the given equation.

Solution:

$\frac{1+tan^2A}{1+cot^2A}$

$=\frac{(1+\frac{sin^2A}{cos2A})}{(1+\frac{cos^2A}{sin^2A})}$

$=\frac{\frac{(cos^2A+sin^2A)}{cos^2A}}{\frac{(sin^2A+cos^2A)}{sin^2A}}$

$=\frac{(\frac{1}{cos^2A})}{(\frac{1}{sin^2A})}$

$=\frac{1}{cos^2A}\times \frac{sin^2A}{1}$

$=\frac{sin^2A}{cos^2A}$

$=tan^2A$

$(\frac{1-tanA}{1-cotA})^2$

$=\frac{(1+tan^2A-2tanA)}{(1+cot^2-2cotA)}$

$=\frac{(sec^2A-\frac{2sinA}{cosA})}{(cosec^2A-\frac{2cosA}{sinA})}$

$=\frac{\frac{1}{cos^2A}-\frac{2sinA}{cosA}}{\frac{1}{sin^2A}-\frac{2cosA}{sinA}}$

$=\frac{\frac{(1-2sinAcosA)}{cos^2A}}{\frac{(1-2cosAsinA)}{sin^2A}}$

$=\frac{(1-2sinAcosA)}{cos^2A} \times \frac{sin^2A}{(1-2sinAcosA)}$

$=\frac{sin^2A}{cos^2A}$

$=tan^2A$

Hence proved. 

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Updated on: 10-Oct-2022

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