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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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Let $a,b,c$ be positive real numbers .Let $\theta=\tan^{-1}\sqrt{\large\frac{a(a+b+c)}{bc}}+$$\tan^{-1}\sqrt{\large\frac{b(a+b+c)}{ca}}$$+\tan^{-1}\sqrt{\large\frac{c(a+b+c)}{ab}}$.Then $\tan\theta$

$(a)\;\pi\qquad(b)\;\large\frac{\pi}{2}$$\qquad(c)\;2\pi\qquad(d)\;None\;of\;these$

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1 Answer

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Step 1:
Let $a+b+c=u$ then
$\theta=\tan^{-1}\sqrt{\large\frac{au}{bc}}$$+\tan^{-1}\sqrt{\large\frac{bu}{ca}}$$+\tan^{-1}\sqrt{\large\frac{cu}{ab}}$
Now we know that
$\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\big(\large\frac{x+y}{1-xy}\big)$ when $xy > 1$
$\sqrt{\large\frac{au}{bc}}\times \sqrt{\large\frac{bu}{ca}}=\large\frac{u}{c}$
$\Rightarrow \large\frac{a+b+c}{c} $$> 1;a,b,c$ being +ve real nos.
Step 2:
We get
$\theta=\pi+\tan^{-1}\bigg[\large\frac{\sqrt{\Large\frac{au}{bc}}+\sqrt{\Large\frac{bu}{ca}}}{1-\sqrt{\Large\frac{au}{bc}}\sqrt{\Large\frac{bu}{ca}}}\bigg]$$+\tan^{-1}\sqrt{\large\frac{cu}{ab}}$
$\theta=\pi+\tan^{-1}\bigg[\large\frac{\Large\frac{a+b}{\sqrt {abc}}\sqrt u}{1-\Large\frac{u}{c}}\bigg]+$$\tan^{-1}\sqrt{\large\frac{cu}{ab}}$
$\theta=\pi+\tan^{-1}\sqrt{\large\frac{uc}{ab}}$$+\tan^{-1}\sqrt{\large\frac{cu}{ab}}$
Using $\tan^{-1}(-x)=-\tan^{-1}x$
$\Rightarrow \pi$
Hence (a) is the correct answer.
answered Oct 21, 2013 by sreemathi.v
 

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