Question

if sin theta = n sin (theta + 2 alpha) then prove that tan ( theta + alpha) =1+n/1-n tan

Answer 1

$\textbf{Given:}$

$sin\,\theta=n\,sin(\theta+2\,\alpha)$

$\textbf{To prove:}$

$tan(\theta+\alpha)=\dfrac{1+n}{1-n}\,tan\alpha$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{1+n}{1-n}\,tan\alpha$

$=\dfrac{1+\frac{sin\,\theta}{sin(\theta+2\,\alpha)}}{1-\frac{sin\,\theta}{sin(\theta+2\,\alpha)}}\,tan\alpha$

$=\dfrac{sin(\theta+2\,\alpha)+sin\,\theta}{sin(\theta+2\,\alpha)-sin\,\theta}\,tan\alpha$

$\text{Using the identities}$

$\boxed{\bf\,sinC+sinD=2\,sin(\frac{C+D}{2})\,cos(\frac{C-D}{2})}$

$\boxed{\bf\,sinC-sinD=2\,cos(\frac{C+D}{2})\,sin(\frac{C-D}{2})}$

$=\dfrac{2\,sin(\theta+\alpha)\,cos\alpha}{2\,cos(\theta+\alpha)\,sin\alpha}\,tan\alpha$

$=\dfrac{sin(\theta+\alpha)\,cos\alpha}{cos(\theta+\alpha)\,sin\alpha}\,\dfrac{sin\alpha}{cos\alpha}$

$=\dfrac{sin(\theta+\alpha)}{cos(\theta+\alpha)}$

$=tan(\theta+\alpha)$

$\implies\boxed{\bf\,tan(\theta+\alpha)=\dfrac{1+n}{1-n}\,tan\alpha}$

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