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Question
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
Solution
LHS = `1/(sin θ + cos θ) + 1/(sin θ - cos θ)`
= `((sin θ - cos θ) + (sin θ + cos θ))/(sin^2 θ - cos^2 θ)`
= `(2 sin θ)/((1 - cos^2 θ) - cos^2 θ)`
= `(2 sin θ)/(1 - 2cos^2 θ)`
= RHS
Hence proved.
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Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
If tan θ = 3, then `(4 sin theta - cos theta)/(4 sin theta + cos theta)` is equal to ______.
(1 – cos2 A) is equal to ______.
