Question
Prove that $\cot ^2 A \left[\frac{\sec A -1}{1+\sin A }\right]+\sec ^2 A \left[\frac{\sin A -1}{1+\sec A }\right]=0$

Answer

L.H.S = `cot^2"A"[(sec"A" - 1)/(1 + sin "A")] + sec^2"A"[(sin"A" - 1)/(1 + sec"A")]`
= `(cot^2"A"(sec"A" - 1)(sec"A" + 1) + sec^2"A"(sin"A" - 1)(sin"A" + 1))/((1 + sin"A")(1 + sec"A"))`
= `(cot^2"A"(sec^2"A" - 1) + sec^2"A"(sin^2"A" - 1))/((1 + sin"A")(1 + sec"A"))`
= `(cot^2"A" xx tan^2"A" + sec^2"A"( - cos^2"A"))/((1 + sin"A")(1 + sec"A"))`
= `(cot^2"A" xx 1/cot^2"A" - sec^2"A" xx 1/sec^2"A")/((1 + sin"A")(1 + sec"A"))`
= `(1 - 1)/((1 + sin"A")(1+ sec"A"))`
= `0/((1 + sin"A")(1 + sec"A"))`
= 0
L.H.S = R.H.S
Hence it is proved.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free