Question

(tan0-cot0)/(sin0×cos0)=tan^20-cot^20​

Answer 1

Step-by-step explanation:

formula used are

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

sec^2θ = 1/ cos^2θ

cosec^2θ = 1/ sin ^2θ

tan^2θ + 1= sec^2θ

cot^2θ + 1 = cosec^2θ

L. H. S = (tanθ -cotθ)/(sinθ ×cosθ)

=( sinθ/cosθ - cosθ/sinθ ) /(sinθ ×cosθ)

= ( sin ^2θ - cos^2θ/sinθcosθ)/(sinθcosθ)

=( sin ^2θ - cos^2θ) / (sin^2θcos^2θ)

= sin ^2θ/sin^2θcos^2θ - cos^2θ/sin^2θcos^2θ

= 1/cos^2θ - 1/sin ^2θ

= sec^2θ - cosec^2θ

= tan^2θ + 1 - (cot^2θ + 1)

= tan^2θ + 1 - cot^2θ - 1

= tan^2θ - cot^2θ

( hence proved)

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Answer 2

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