Question

...don't spam relevant ans only​

Answer 1

question..

( 1+tan2A/1+cos²A) =(1-tanA/1-cotA)² =tan²A

answer.....

★Given that: 1+tan²A1+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★

★hope.it.helps....★

thanks....★