Question

1+tan^2A/1+cot^2A=(1-tanA/1-cotA)^2​

Answer 1

Step-by-step explanation:

LHS

1+tan²A/1+cot²A

=sec²A/cosec²A

=sin²A/cos²A

=tan²A

RHS

(1-tanA/1-cotA)²

={(cosA-sinA/cosA)/(sinA-cosA/sinA)}²

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

={(cosA-sinA)²/cos²A}×{sin²A/(cosA-sinA)²}

=sin²A/cos²A

=tan²A

LHS=RHS

Answer 2

Solution →

Answer is in the attachment.

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