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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Solve $3\tan(\theta-15^{\large\circ})=\tan(\theta+15^{\large\circ})$

$\begin{array}{1 1}(A)\;\large\frac{n\pi}{2}\normalsize +(-1)^n\large\frac{\pi}{4}&(B)\;n\pi\pm \large\frac{\pi}{6}\\(C)\;n\pi\pm \large\frac{\pi}{8}&(D)\;2n\pi\pm \large\frac{3\pi}{4}\end{array} $

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1 Answer

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Toolbox:
  • $\tan\theta=\large\frac{\sin \theta}{\cos \theta}$
  • $\sin(a+b)=\sin a\cos b+\cos a\sin b$
  • $\sin(a-b)=\sin a\cos b-\cos a\sin b$
Given
$3\tan (\theta-15^{\large\circ})=\tan (\theta+15^{\large\circ})$
$\large\frac{\tan (\theta+15^{\large\circ})}{\tan (\theta-15^{\large\circ})}=\frac{3}{1}$
By componendo & dividendo
$\large\frac{\tan (\theta+15^{\large\circ})+\tan (\theta-15^{\large\circ})}{\tan (\theta+15^{\large\circ})+\tan (\theta-15^{\large\circ})}=\frac{3+1}{3-1}$
$\Rightarrow \large\frac{4}{2}$$=2$
$\large\frac{\Large\frac{\sin (\theta+15^{\large\circ})}{\cos (\theta+15^{\large\circ})}+\Large\frac{\sin (\theta-15^{\large\circ})}{\cos (\theta-15^{\large\circ})}}{\Large\frac{\sin (\theta+15^{\large\circ})}{\cos (\theta+15^{\large\circ})}-{\Large\frac{\sin (\theta-15^{\large\circ})}{\cos (\theta-15^{\large\circ})}}}=$$2$
$\large\frac{\sin(\theta+15^{\large\circ})\cos(\theta-15^{\large\circ})+\sin(\theta-15^{\large\circ})\cos(\theta+15^{\large\circ})}{\sin(\theta+15^{\large\circ})\cos(\theta-15^{\large\circ})-\sin(\theta-15^{\large\circ})\cos(\theta+15^{\large\circ})}$$=2$
$\large\frac{\sin(\theta+15^{\large\circ}+\theta-15^{\large\circ})}{\sin(\theta+15^{\large\circ}-\theta+15^{\large\circ})}$$=2$
$\Rightarrow \large\frac{\sin 2\theta}{\sin 3\theta}=$$2$
$\Rightarrow 2\sin 2\theta=2$
$\sin 2\theta=1=\sin\large\frac{\pi}{2}$
$\therefore 2\theta=n\pi+(-1)^n\large\frac{\pi}{2}$
$\therefore \theta=\large\frac{n\pi}{2}$$+(-1)^n\large\frac{\pi}{4}$
Hence (A) is the correct answer.
answered May 22, 2014 by sreemathi.v
 

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