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Question
Prove that `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1) = (1 + cos "A")/sin "A"`
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Solution
LHS = `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1)`
= `((cot "A" + "cosec A") - ("cosec"^2 "A" - cot^2 "A"))/(cot "A" - "cosec A" + 1)`
= `((cot "A" + "cosec A")("cosec A" + cot "A")("cosec A" - cot "A"))/(cot "A" - "cosec A" + 1)`
= `((cot "A" + "cosec A") [1 - "cosec A" - cot "A"])/(cot"A"-"cosec A"+1)`
= `((cot "A" + "cosec A") (1-"cosec A"+cot"A"))/(1-"cosec A"+cot"A")`
= cot A + cosec A
= `cos"A"/sin"A"+1/sin"A"=(cos"A"+1)/sin"A"`
= `(1+cos"A")/sin"A"`
= RHS
Hence proved.
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