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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`
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L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
