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Question
Prove the following identities:
(1 + cot A – cosec A)(1 + tan A + sec A) = 2
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Solution
L.H.S. = (1 + cot A – cosec A)(1 + tan A + sec A)
= `(1 + cosA/sinA - 1/sinA)(1 + sinA/cosA + 1/cosA)`
= `((sinA + cosA - 1)/sinA)((cosA + sinA + 1)/cosA)`
= `((sinA + cosA - 1)(sinA + cosA + 1))/(sinAcosA)`
= `((sinA + cosA)^2 - (1)^2)/(sinAcosA)`
= `(sin^2A + cos^2A + 2sinAcosA - 1)/(sinAcosA)`
= `(1 + 2sinAcosA - 1)/(sinAcosA)`
= `(2sinAcosA)/(sinAcosA)`
= 2 = R.H.S.
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Show that, cotθ + tanθ = cosecθ × secθ
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
