Inverse Circular Functions 3 Question 11
11. 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: 11. 0
Solution:
- Given,
$$ \begin{array}{r} \theta=\tan ^{-1} \sqrt{\frac{a(a+b+c)}{b c}}+\tan ^{-1} \sqrt{\frac{b(a+b+c)}{a c}} \\ +\tan ^{-1} \sqrt{\frac{c(a+b+c)}{a b}} \\ \because \tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\tan ^{-1} \frac{x+y+z-x y z}{1-x y-y z-z x} \end{array} $$
$$ =\tan ^{-1} \frac{-(a+b+c) \sqrt{\frac{a+b+c}{a b c}}}{1-(a+b+c) \frac{1}{a}+\frac{1}{b}+\frac{1}{c}} $$
$$ \begin{aligned} & =\tan ^{-1} \frac{\sqrt{\frac{a+b+c}{a b c}}(a+b+c)-(a+b+c) \sqrt{\frac{a+b+c}{a b c}}}{1-\frac{(a+b+c)(a b+b c+c a)}{a b c}} \\ & \Rightarrow \quad \theta=\tan ^{-1} 0 \Rightarrow \tan \theta=0 \end{aligned} $$
