If an angle $θ$ be divided into two parts such that the tangent of one part is $m$ times the tangent of other, then prove that their difference $ϕ$ is obtained from the equation $sin ϕ = [(m - 1) / (m = 1)] sin θ$

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Solution

Let the two parts be $A$ And $B$ so that
$A + B = θ; A - B = ϕ; tan A = m tan B$
$∴ \frac{tan A}{tan B} = \frac{m}{1}$
Apply componendo and dividendo
$∴ \frac{tan A - tan B}{tan A + tan B} = \frac{m - 1}{m + 1}$
or $\frac{sin (A - B)}{sin (A + B)} = \frac{m - 1}{m + 1}$
or $sin ϕ = {(m - 1) / (m + 1)} sin θ$
Hence proved.

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If an angle θ be divided into two parts such that the tangent of one part is m times the tangent of other, then prove that their difference ϕ is obtained from the equation sinϕ=[(m−1)/(m=1)]sinθ

Let the two parts be A And B so thatA+B=θ;A−B=ϕ;tanA=mtanB∴tanAtanB=m1Apply componendo and dividendo∴tanA−tanBtanA+tanB=m−1m+1or sin(A−B)sin(A+B)=m−1m+1or sinϕ={(m−1)/(m+1)}sinθHence proved.

If an angle $θ$ be divided into two parts such that the tangent of one part is $m$ times the tangent of other, then prove that their difference $ϕ$ is obtained from the equation $sin ϕ = [(m - 1) / (m = 1)] sin θ$