Question

tan^2x+ cot^2x+2 = sec^2 × cosec^2



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Answer 1

tan^2x+ cot^2x+2 = sec^2 × cosec^2

sin^2x/cos^2x + cos^2x/sin^2x + 2 = 1/cos^2x × 1/sin^2x

(sin^4x + cos^4x)/cos^2x sin^2x + 2sin^2 xcos^2x/ cos^2x sin^2x = 1/sin^2x cos^2x

(sin^4x + cos^4x)/cos^2x sin^2x + 2sin^2 xcos^2x/ cos^2x sin^2x

this is the identity

(a+b)^2

(sin^2x + cos^2x)^2

sin^2x + cos^2x = 1

so, 1/ cos^2x sin^2x = 1/ cos^2x sin^2x

LHS = RHS

Answer 2

plz their are not 2

this equation is

tan'2x+cot'2x=sec'2x*cosec'2x

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I hope that you got answer