Question

6. Prove that (1 + cotA + tana) (sinA - COSA) =secA/cesec2A-cosecA/sec2A​

Answer 1

Answer:

L.H.S.

(1+cotA+tanA)(sinA−cosA)

= (1+cosA/sinA+sinA/cosA)(sinA−cosA)

= (sinA−cosA)(sinAcosA+sin2A+cos2A)/sinAcosA

= sin3A−cos3A/sinAcosA

= sin2A/cosA−cos2/AsinA

= 1/cosA.sin2A/1−1/sinA.cos2A/1

= secA/cosec2A−cosecA/sec2A

= R.H.S.