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Question
Prove the following identities:
`(secA - tanA)/(secA + tanA) = 1 - 2secAtanA + 2tan^2A`
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Solution
L.H.S. = `(secA - tanA)/(secA + tanA)`
= `(secA - tanA)/(secA + tanA) xx (secA - tanA)/(secA - tanA)`
= `(secA - tanA)^2/(sec^2A - tan^2A)`
= `(sec^2A + tan^2A - 2secAtanA)/1`
= 1 + tan2 A + tan2 A – 2 sec A tan A
= 1 – 2 sec A tan A + 2 tan2 A = R.H.S.
RELATED QUESTIONS
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(i) (1 – sin2θ) sec2θ = 1
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`cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A`
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cos 1°cos 2°cos 3° ....cos 180° = 0.
Prove that: `1/(cosec"A" - cot"A") - 1/sin"A" = 1/sin"A" - 1/(cosec"A" + cot"A")`
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
