Question

Show that :

1-tan² A / cot² A - 1 = tan² A

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Answer 1

Given:

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

To Prove:

LHS = RHS

Solution:

We know that ,

$\tan \theta = \dfrac{1}{\cos \theta}$

$\dfrac{1 - { \tan}^{2}A }{ { \cot}^{2}A - 1 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \dfrac{1 - { \tan}^{2}A }{ \dfrac{1}{ { \tan }^{2}A } - 1 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \dfrac{1 - { \tan}^{2} A}{ \dfrac{1 - { \tan}^{2} A}{ { \tan}^{2}A } } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \dfrac{1 - { \tan}^{2}A }{1 - { \tan }^{2}A } \times { \tan}^{2} A \\ \\ { \tan }^{2} A \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \mathfrak{hence \: proved} \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Hope it helps u

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Answer 2

Answer:

Hey Brainly User !

_______________________________________

As we know that,cotA = 1/tanA. _______________________________________

L.H.S = (1 - tan²A)/(cot²A - 1) = tan²A

= (1 - tan²A)/[(1/tan²A - 1/1)] = tan²A

= (1 - tan²A)/[(1/tan²A)]/tan²A)]

After Cancellation,

tan²A = tan²A

L.H.S = R.H.S

Hence Proved.

I hope it helps.

:)