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प्रश्न
Prove that `"cot A"/(1 - cot"A") + "tan A"/(1 - tan "A")` = – 1
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उत्तर
L.H.S = `"cot A"/(1 - cot"A") + "tan A"/(1 - tan "A")`
= `"cot A"/(1 - 1/(tan"A")) + "tan A"/(1 - tan "A")`
= `"cot A"/((tan "A" - 1)/(tan "A")) + "tan A"/(1 - tan "A")`
= `"cot A tan A"/(tan "A" - 1) + "tan A"/(1 - tan "A")`
= `1/(tan "A" - 1) + "tan A"/(1 - tan "A")` ......[∵ cot A tan A = 1]
= `- 1/(1 - tan "A") + "tan A"/(1 - tan "A")`
= `- (1/(1 -tan "A") - "tan A"/(1- tan "A"))`
= `-((1 - tan "A")/(1 - tan "A"))`
= – 1
= R.H.S
∴ `"cot A"/(1 - cot"A") + "tan A"/(1 - tan "A")` = – 1
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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
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
