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Prove the following : $ 2tan^{-1} \bigg(\large \frac{1}{2} \bigg)$$ +\tan^{-1} \bigg( \large\frac{1}{7} \bigg)$$ = \tan^{-1} \bigg( \large\frac{31}{17} \bigg) $

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1 Answer

Toolbox:
  • $ 2\tan^{-1}x=\tan^{-1}\large\frac{2x}{1-x^2}$
  • $\tan^{-1}x+tan^{-1}y=\tan^{-1}\large\frac{x+y}{1-xy}$
Given :
$2\tan^{-1}\big(\large\frac{1}{2}\big)$$+\tan^{-1}\big(\large\frac{1}{7}\big)$
(i.e) $\big[\tan^{-1}\big(\large\frac{1}{2}\big)$$+\tan^{-1}\big(\large\frac{1}{2}\big)\big]$$+\tan^{-1}\big(\large\frac{1}{7}\big)$
$\Rightarrow \big[\tan^{-1}\bigg(\large\frac{1/2+1/2}{1-1/2\times 1/2}\big]+\tan^{-1}\big(\large\frac{1}{7}\big)$
$\Rightarrow \tan^{-1}\big[\large\frac{1}{1-\Large\frac{1}{4}}\big]+$$\tan^{-1}\big(\large\frac{1}{7}\big)$
$\Rightarrow \tan^{-1}\big[\large\frac{1}{\Large\frac{3}{4}}\big]$$+\tan^{-1}\big(\large\frac{1}{7}\big)$
$\Rightarrow \tan^{-1}\big(\large\frac{4}{3}\big)$$+\tan^{-1}\big(\large\frac{1}{7}\big)$
$\tan^{-1}\bigg[\large\frac{\Large\frac{28+3}{21}}{\Large\frac{17}{21}}\big]$
$\tan^{-1}\big[\large\frac{31}{17}\big]$
LHS=RHS
Hence proved.
answered Nov 14, 2013 by sreemathi.v
 
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