Question

Heyyy guys....
answer this one!!!
#HAVE A LOOK AT THE ATTACHMENT...!!! ​

Answer 1

Answer:

Proved below.

Step-by-step-explanation:

$\implies \bigg( \dfrac{1 + \tan {}^{2} ( \alpha ) }{1 + \cot {}^{2} ( \alpha ) } \bigg) \\ \\ \\ \implies \bigg( \dfrac{1 + \tan {}^{2} ( \alpha ) }{1 + \frac{1}{ \tan {}^{2} ( \alpha ) } } \bigg) \\ \\ \\ \implies \bigg( \dfrac{1 + { \tan {}^{2} ( \alpha ) }^{} }{ \frac{\tan {}^{2} ( \alpha ) + 1}{ \tan {}^{2} ( \alpha ) } } \bigg) \\ \\ \\ \implies \tan {}^{2} ( \alpha )$

$\implies ( \tan( \alpha ) ) {}^{2} \\ \\ \implies \bigg( \dfrac{1}{ \frac{1}{ \tan {}^{} ( \alpha ) } } \bigg) {}^{2} \\ \\ \\ \implies \bigg( \dfrac{1 - \tan( \alpha ) }{ \frac{1 - \tan( \alpha ) }{ \tan( \alpha ) } } \bigg) {}^{2} \\ \\ \\ \implies \bigg( \dfrac{1 - \tan( \alpha ) }{ \frac{1}{ \tan( \alpha ) } - \frac{ \tan( \alpha ) }{ \tan( \alpha ) } } \bigg) {}^{2} \\ \\ \\ \implies \bigg(\dfrac{1 - \tan( \alpha ) }{ \cot( \alpha ) - 1} \bigg) {}^{2} \\ \\ \\ \implies \bigg( \dfrac{1 - \tan( \alpha ) }{1 - \cot( \alpha ) } \bigg) {}^{2}$

Hence proved.

Answer 2

$\huge\underline\mathfrak\pink{Answer:-}$

(-sinA/cosA)^2 = (-tanA)^2 =tan^2 A.

STEP BY STEP EXPLANATION:-

(1+tan^2A)/(1+cot^2A) = sec^2A/cosec^2A

=(1/cos^2A)/(1/sin^2A)=sin^2A/cos^2A

= tan^2A.

Again {(1-tanA)/(1-cotA)}^2

=[{1-(sinA/cosA)}/{1-(cosA/sinA)}]^2

=[{(cosA-sinA)/cosA}/

{(sinA-cosA)/sinA}]^2

={sinA(cosA-sinA)/cosA(sinA-cosA)}^2

={-sinA(sinA-cosA)/cosA(sinA-cosA)}^2

=(-sinA/cosA)^2 = (-tanA)^2 =tan^2 A.

______________________

Hope it is clear!