Question

(1+cotA+tanA)(sinA-cosA)=(secA/cosec^2A)-(cosecA/sec^2A)=sinAtanA-cotAcosA
plz solve it step by step....​

Answer 1

♠♦♠ FOLLOW♠♦♠

Step-by-step explanation:

1+cotA+tanA)(sinA−cosA)(1+cot⁡A+tan⁡A)(sin⁡A−cos⁡A)

= (1+cosAsinA+sinAcosA)(sinA−cosA)(1+cos⁡Asin⁡A+sin⁡Acos⁡A)(sin⁡A−cos⁡A)

= (sinA−cosA)(sinAcosA+sin2A+cos2A)sinAcosA(sin⁡A−cos⁡A)(sin⁡AcosA+sin2A+cos2A)sin⁡AcosA

= sin3A−cos3AsinAcosAsin3A−cos3Asin⁡AcosA

= sin2AcosA−cos2AsinAsin2AcosA−cos2Asin⁡A

= 1cosA.sin2A1−1sinA.cos2A11cosA.sin2A1−1sinA.cos2A1

= secAcosec2A−cosecAsec2Asec⁡Acos⁡ec2A−cos⁡ecAsec2A

= R.H.S.