Question

1-sin2A/cos2A=1-tanA/1+tanA

Answer 1

we have , sin 2A = 2 sinA cosA , cos2A = cos ^2 A - sin ^2 A 
and sin ^2 A + cos ^2 A = 1 
[( cos ^2 A + sin ^2 A - 2sin A cos A ) / ( cos^2 A - sin^2 A) ] = 
[ 1 - ( sin A / cos A)] / [ 1 + ( sinA / cos A)] 
[ ( cos A - sin A) ^2] / [(cosA - sinA)( cos A + sinA)] = [(cosA - sin A) / cos A] / [ (cos A + sin A ) / cosA] 

[(cos A - sin A) / ( cos A + sin A )] = [( cos A - sinA) / cos A] [ cos A / (cos A + sin A) ]

Answer 2

Step-by-step explanation:

R•H•S.

1-tanA/1+tanA

=1-sinA/cosA/1+sinA/cosA

=cosA-sinA/cosA/cosA+sinA/cosA

=cosA-sinA/cosA+sinA

=प्रमेयी करण करने पर cosA-sinA

=(cosA-sinA)×(cosA-sinA)/(cosA+sinA)×(cosA-sinA)

(cosA-sinA)^2)/(cos^2-sin^2A)

=

=cos^2+sin^2A-2sinA.cosA/cos2A

=1-2sinA.cosA/cos2A

=1-sin2A/cos2A