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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Prove that $\tan(60^{\large\circ}+A)\tan(60^{\large\circ}-A)=\large\frac{2\cos 2A+1}{2\cos 2A-1}$

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Toolbox:
  • $2\cos A.\cos B=\cos(A+B)+\cos (A-B)$
  • $2\sin A.\sin B=\cos(A-B)-\cos (A+B)$
$\tan (60^{\large\circ}+A).\tan (60^{\large\circ}-A)$
$\Rightarrow \large\frac{\sin (60^{\large\circ}+A).\sin (60^{\large\circ}-A)}{\cos (60^{\large\circ}+A).\cos (60^{\large\circ}-A)}$
$\Rightarrow \large\frac{2\sin (60^{\large\circ}+A).\sin (60^{\large\circ}-A)}{2\cos (60^{\large\circ}+A).\cos (60^{\large\circ}-A)}$
$\Rightarrow \large\frac{\cos[ (60^{\large\circ}+A)-(60^{\large\circ}-A)]-\cos [ (60^{\large\circ}+A)-(60^{\large\circ}-A)]}{\cos[ (60^{\large\circ}+A)-(60^{\large\circ}-A)]+\cos [ (60^{\large\circ}+A)-(60^{\large\circ}-A)]}$
$\Rightarrow \large\frac{\cos 2A-\cos 120}{\cos 120+\cos 2A}$
$\Rightarrow \large\frac{\cos 2A-(-\Large\frac{1}{2})}{(-\Large\frac{1}{2})\normalsize +\cos 2A}$
$\Rightarrow \large\frac{2\cos 2A+1}{2\cos 2A-1}$
Hence proved
answered May 21, 2014 by sreemathi.v
 

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