Question

prove that (cot^2(90-theta)/tan^2theta-1)+(cosec^2theta/(sec^2 theta - cosec^2 theta))= (1/(sin^2 theta - cos^2theta))​

Answer 1

Answer:

We know that,

cot2 A = cos2 A/ sin2 A and tan2 A = sin2 A/cos2 A

Substituting the above in L.H.S, we get

L.H.S = sin2 A cot2 A + cos2 A tan2 A

= {sin2 A (cos2 A/ sin2 A)} + {cos2 A (sin2 A/cos2 A)}

= cos2 A + sin2 A

= 1 [∵ sin2 θ + cos2 θ = 1]

= R.H.S

– Hence Proved

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Answer 2

Step-by-step explanation:

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