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Question
If m sinθ = n sin(θ + 2α), then prove that tan(θ + α)cotα = `(m + n)/(m - n)`
[Hint: Express `(sin(theta + 2alpha))/sintheta = m/n` and apply componendo and dividendo]
Solution
Given that: m sinθ = n sin(θ + 2α)
⇒ `(sin(theta + 2alpha))/sintheta = m/n`
Using componendo and dividendo theorem, we get,
⇒ `(sin(theta + 2alpha) + sintheta)/(sin(theta + 2alpha) - sintheta) = (m + n)/(m - n)`
⇒ `(2sin((theta + 2alpha + theta)/2).cos((theta + 2alpha - theta)/2))/(2cos((theta + 2alpha + theta)/2).sin((theta + 2alpha - theta)/2)) = (m + n)/(m - n)` .......`[(because sinA + sinB = 2sin (A + B)/2 . cos (A - B)/2),(sinA - sinB = 2cos (A + B)/2 . sin (A - B)/2)]`
⇒ `(sin(theta + alpha).cosalpha)/(cos(theta + alpha).sinalpha) = (m + n)/(m - n)`
⇒ `tan(theta + alpha)cotalpha = (m + n)/(m - n)`
Hence proved.
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