Solve - 1+tan2 A\1+cot2A=[1-tanA\1-cotA]2=tan2A.

Given that: 1+tan² A1+cot²A=[1-tanA1-cotA]²=tan²A

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.

=tan

The values of LHS and RHS are same.

Hence proved

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