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Question
Prove that
`cot^2A-cot^2B=(cos^2A-cos^2B)/(sin^2Asin^2B)=cosec^2A-cosec^2B`
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Solution
`cot^2A-cot^2B`
`=cos^2A/sin^2A-cos^2B/sin^2B`
`=(cos^2Asin^2B-cos^2Bsin^2A)/(sin^2Asin^2B)`
`=(cos^2A(1-cos^2B)-cos^2B(1-cos^2A))/(sin^2Asin^2B)`
`=(cos^2A-cos^2Acos^2B-cos^2B+cos^2Bcos^2A)/(sin^2Asin^2B)`
`=(cos^2A-cos^2B)/(sin^2Asin^2B)`
`=(1-sin^2A-1+sin^2B)/(sin^2Asin^2B)`
`=(-sin^2A+sin^2B)/(sin^2Asin^2B)`
`=sin^2B/(sin^2AsinB)-sin^2A/(sin^2Asin^2B)`
`=1/sin^2A-1/sin^2B`
= cosec2A - cosec2B
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