Question

2
1 + tan^2A
1 + cot^2A
=
1 - tan A
1 - cot A
=tanA​

Answer 1

Step-by-step explanation:

We will first solve the equation on LHS

LHS:

= 1+tan²A / 1+cot²A

Using the trignometric identities we know that 1+tan²A= Sec²A and 1+cot²A= Cosec²A

= Sec²A/ Cosec²A

On taking the reciprocals we get

= Sin²A/Cos²A

= tan²A

RHS:

=(1-tanA)²/(1-cotA)²

Substituting the reciprocal value of tan A and cot A we get,

=(1-sinA/cosA)²/(1-cosA/sinA)²

=[(cosA-sinA)/cosA]²/ [(sinA-cos)/sinA)²

=(cosA-sinA)²×sin²A /Cos²A. /(sinA-cosA)²

=1×sin²A/Cos²A×1.

Answer 2

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