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प्रश्न
Prove that : `tan"A"/(1 - cot"A") + cot"A"/(1 - tan"A") = sec"A".cosec"A" + 1`.
उत्तर
LHS = `(sin"A"/cos"A")/(1 - cos"A"/sin"A") + (cos"A"/sin"A")/(1 - sin"A"/cos"A")`
= `(sin"A" sin"A")/(cos"A"(sin"A" - cos"A")) + (cos"A" cos"A")/((cos"A" - sin"A") sin"A"`
= `1/((sin"A" - cos"A")) [(sin^2"A")/cos"A" + (cos^2"A")/(-sin"A")]`
= `(sin^3"A" - cos^3"A")/(sin"A".cos"A"(sin"A" - cos"A"))`
= `((sin"A" - cos"A")(sin^2"A" + cos^2"A" + sin"A". cos"A"))/(sin"A". cos"A"(sin"A" - cos"A")`
= `(1 + sin"A". cos"A")/(sin"A".cos"A")`
= `1/(sin"A".cos"A") + (sin"A".cos"A")/(sin"A".cos"A")`
= `1/sin"A" . 1/cos"A"+ 1`
= sec A.cosec A + 1
= RHS
Hence proved.
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