Question

PT (1+sin2A)/cos2A = (cosA+sinA)/cosA-sinA​

Answer 1

Step-by-step explanation:

cosA1−tanA+sin2AsinA−cosALHS=cos⁡A1−tan⁡A+sin2⁡Asin⁡A−cos⁡A

=cosA1−sinAcosA+sin2AsinA−cosA=cos⁡A1−sin⁡Acos⁡A+sin2⁡Asin⁡A−cos⁡A

=cos2AcosA−sinA+sin2AsinA−cosA=cos2⁡Acos⁡A−sin⁡A+sin2⁡Asin⁡A−cos⁡A

=−cos2AsinA−cosA+sin2AsinA−cosA=−cos2⁡Asin⁡A−cos⁡A+sin2⁡Asin⁡A−cos⁡A

=−cos2A+sin2AsinA−cosA=−cos2⁡A+sin2⁡Asin⁡A−cos⁡A

=sin2A−cos2AsinA−cosA=sin2⁡A−cos2⁡Asin⁡A−cos⁡A

=(sinA+cosA)(sinA−cosA)sinA−cosA