Question

Q.33. √sec2 + cosec2= tane + cote
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Answer 1

Consider LHS: 2sec2θ-sec4θ-2cosec2θ+cosec4θ = 2sec2θ – 2cosec2θ – sec4θ + cosec4θ = 2(sec2θ – cosec2θ) – (sec4θ – cosec4θ) = 2(sec2θ – cosec2θ) – (sec2θ – cosec2θ) (sec2θ + cosec2θ) = (sec2θ – cosec2θ)[2 – (sec2θ + cosec2θ)] = (sec2θ – cosec2θ)[2 – (1 + tan2θ + 1 + cot2θ)] = (1 + tan2θ – 1 – cot2θ)[2 – (2 + tan2θ + cot2θ)] = (tan2θ – cot2θ) [2 – 2 – tan2θ – cot2θ)] = (tan2θ – cot2θ) × – [tan2θ + cot2θ)] = (cot2θ – tan2θ) × [cot2θ + tan2θ)] = cot4θ – tan4θ = RHS❤❤❤ SINGHBOY ❤❤