Question

Prove that: (cot A + sec B)2 – (tan B – cosec A)2 = 2(cot A . sec B + tan B. cosec A) . Hi, Answer this​

Answer 1

Answer:

LHS = (cotA + secB)^2 - (tanB - cosecA)2

by using (a+b)^2= a^2 + 2ab + b^2

&. (a-b)^2 = a^2 - 2ab + b^2

= cot^A + 2.cotA.secB + sec^2B - (tan^2B -

2tanB.cosecB + cosec^2A)

= cot^2A + 2cotA.secB + sec^2B - tan^2B +

2tanB.cosecB - cosec^2A

using identities cot^2A = cosec^2A - 1

&. sec^2A = tan^2A +1

= cosec^2A - 1 + 2cotA.secB + tan^2A +1 - tan^2B +

2tanB.cosecB - cosec^2A

= 2cotA.secB + 2tanB.cosecB

= 2 (cotA.secB + tanB.cosecB)

= RHS

hence proved

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