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Question
Prove the following identities:
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
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Solution
L.H.S. = `1/(sinA + cosA) + 1/(sinA - cosA)`
= `(sinA - cosA + sinA + cosA)/((sinA + cosA)(sinA - cosA))`
= `(2sinA)/(sin^2A - cos^2A)`
= `(2sinA)/(1 - cos^2A - cos^2A)` ...(∵ sin2A = 1 – cos2A)
= `(2sinA)/(1 - 2cos^2A)`
= R.H.S.
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RELATED QUESTIONS
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To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
