Inverse Circular Functions Ques 20
- If $a, b, c$ are positive real numbers $\theta=\tan ^{-1} \sqrt{\frac{a(a+b+c)}{b c}}+\tan ^{-1} \sqrt{\frac{b(a+b+c)}{c a}}$
$ +\tan ^{-1} \sqrt{\frac{c(a+b+c)}{a b}} $
Then, $\tan \theta$ equals
(1981, 2M)
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Answer:
Correct Answer: 20.0
Solution:
Formula:
Identities of Addition and Substraction:
- Given,
$ \begin{array}{r} \theta=\tan ^{-1} \sqrt{\frac{a(a+b+c)}{b c}}+\tan ^{-1} \sqrt{\frac{b(a+b+c)}{a c}} \\ +\tan ^{-1} \sqrt{\frac{c(a+b+c)}{a b}} \\ [\because \tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\tan ^{-1} (\frac{x+y+z-x y z}{1-x y-y z-z x})] \end{array} $
$ =\tan ^{-1} [\frac{-(a+b+c) \sqrt{\frac{a+b+c}{a b c}}}{1-(a+b+c) \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}] $
$ \begin{aligned} & =\tan ^{-1} [\frac{\sqrt{\frac{a+b+c}{a b c}}(a+b+c)-(a+b+c) \sqrt{\frac{a+b+c}{a b c}}}{1-\frac{(a+b+c)(a b+b c+c a)}{a b c}}] \\ & \Rightarrow \quad \theta=\tan ^{-1} 0 \Rightarrow \tan \theta=0 \end{aligned} $
BETA
