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प्रश्न
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
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उत्तर
To Prove: `((1 + tan^2 A))/((1 + cot^2 A))` = sec2 A – 1
LHS.
We have, `(((1 + sin^2 A)/(cos^2 A)))/(((1 + cos^2 A)/(sin^2 A)))`
= `[(((cos^2 A + sin^2 A))/(cos^2 A))/(((sin^2 A + cos^2 A))/(sin^2 A))]`
= `((1/cos^2 A))/((1/sin^2 A))` ...[As sin2 A + cos2 A = 1]
= `((sin^2 A))/((cos^2 A))`
= tan2 A
= sec2 A – 1
Hence, proved.
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संबंधित प्रश्न
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Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
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∴ cotθ + tanθ = cosecθ × secθ
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