Question

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Evaluate : 1+tan²A/1+cot²A​

Answer 1

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Given that: 1+ tan² A/1+ cot²A = [1 – tanA/1- 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

LHS = 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.

= tan²A

The values of LHS and RHS are the same.

Hence proved

Answer 2

Answer:

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Step-by-step explanation:

Given that: 1+ tan² A/1+ cot²A = [1 – tanA/1- 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

LHS = 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.

= tan²A

The values of LHS and RHS are the same.

Hence proved