Question

Please solve this question ??​

Answer 1

Given:

Proof:

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

Solution:

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

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

On R.H.S :

Using ID: $\rm (a - b)^2 = a^2 - 2ab + b^2$

$\therefore \rm \frac{[(1)^2 - 2\times 1 \times \tan A + (\tan A)^2]}{[(1)^2 - 2\times 1 \times \cot A + (\cot A)^2]}$

$\implies \rm \bf \frac{[1 - 2\tan A + \tan^2 A]}{[1 - 2\cot A + \cot^2 A]}$

On L.H.S :

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

$\implies \rm \frac{(1 + \tan A)^2}{(1 + \cot A)^2}$

$\implies \bf \rm \frac{[1 + 2 \tan A + \tan^2 A]}{[1 + 2 \cot A + \cot^2 A]}$

Therefore,

$\therefore \rm \frac{[1 + 2 \tan A + \tan^2 A]}{[1 + 2 \cot A + \cot^2 A]} = \frac{[1 - 2\tan A + \tan^2 A]}{[1 - 2\cot A + \cot^2 A]}$

Cancelling from both sides,

$\implies\bf \rm \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{1 + \tan^2 A}{1 + \cot^2 A}$

L.H.S = R.H.S

HENCE, PROVED.