Question

prove the question given in attachment​

Answer 1

Step-by-step explanation:

I didn't have time to write but here's your answers

Answer 2

Solution!!

$\sf \dfrac{1+\tan ^{2}A}{1+\cot ^{2}A}=\dfrac{1-\tan ^{2}A}{1-\cot ^{2}A}=\tan ^{2}A$

Taking LHS,

$\sf =\dfrac{1+\tan ^{2}A}{1+\cot ^{2}A}$

$\sf =\dfrac{1+\tan ^{2}A}{1+\dfrac{1}{\tan ^{2}A}}$

$\sf =\dfrac{1+\tan ^{2}A}{\dfrac{\tan ^{2}A+1}{\tan ^{2}A}}$

$\sf =\dfrac{1+\tan ^{2}A}{\tan ^{2}A+1}\times \tan ^{2}A$

$\sf =\tan^{2}A$

LHS = RHS

Taking Middle Term,

$\sf =\left(\dfrac{1-\tan A}{1-\cot A}\right)^{2}$

$\sf =\left(\dfrac{1-\tan A}{1-\dfrac{1}{\tan A}}\right)^{2}$

$\sf =\left(\dfrac{1-\tan A}{\dfrac{\tan A-1}{\tan A}}\right)^{2}$

$\sf =\left(-\tan A\right)^{2}$

$\sf =\tan ^{2}A$

Middle term = RHS

LHS = Middle Term = RHS

Hence, proved.