Question

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If tanθ = Q sinα / ( P + Q cosα ) , Then Prove that ,

tan ( α - θ ) = P sinα / ( Q + P cosα )


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Answer 1

$\underline{\textsf{Proof :}}$

$\mathsf{Given\:that,\:tan\theta=\frac{Qsin\alpha}{P+Qcos\alpha}}$

$\mathsf{Then,\:tan(\alpha-\theta)}$

$\mathsf{=\frac{tan\alpha-tan\theta}{1+tan\alpha\:tan\theta}}$

$\mathsf{=\dfrac{\frac{sin\alpha}{cos\alpha}-\frac{Qsin\alpha}{P+Qcos\alpha}}{1+\frac{sin\alpha}{cos\alpha}\frac{Qsin\alpha}{P+Qcos\alpha}}}$

$\mathsf{=\frac{sin\alpha(P+Qcos\alpha)-Qsin\alpha\:cos\alpha}{Pcos\alpha+Qcos^{2}\alpha+Qsin^{2}\alpha}}$

$\mathsf{=\frac{Psin\alpha+Qsin\alpha\:cos\alpha-Qsin\alpha\:cos\alpha}{(sin^{2}\alpha+cos^{2}\alpha)Q+Pcos\alpha}}$

$\mathsf{=\frac{Psin\alpha}{Q+Pcos\alpha}}$

$\implies \boxed{\mathsf{tan(\alpha-\theta)=\frac{Psin\alpha}{Q+Pcos\alpha}}}$

$\textsf{Hence, proved.}$

Answer 2

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