SATHEE: Inverse Circular Functions 3 Question 11

Inverse Circular Functions 3 Question 11

11. If $a, b, c$ are positive real numbers $θ = \tan^{- 1} \sqrt{\frac{a (a + b + c)}{b c}} + \tan^{- 1} \sqrt{\frac{b (a + b + c)}{c a}}$

$+ \tan^{- 1} \sqrt{\frac{c (a + b + c)}{a b}}$

Then, $\tan θ$ equals

(1981, 2M)

Show Answer

Answer:

Correct Answer: 11. 0

Solution:

Given,

$\begin{array}{r} θ = \tan^{- 1} \sqrt{\frac{a (a + b + c)}{b c}} + \tan^{- 1} \sqrt{\frac{b (a + b + c)}{a c}} \\ + \tan^{- 1} \sqrt{\frac{c (a + b + c)}{a b}} \\ ∵ \tan^{- 1} x + \tan^{- 1} y + \tan^{- 1} z = \tan^{- 1} \frac{x + y + z - x y z}{1 - x y - y z - z x} \end{array}$

$= \tan^{- 1} \frac{- (a + b + c) \sqrt{\frac{a + b + c}{a b c}}}{1 - (a + b + c) \frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$

$\begin{aligned} = \tan^{- 1} \frac{\sqrt{\frac{a + b + c}{a b c}} (a + b + c) - (a + b + c) \sqrt{\frac{a + b + c}{a b c}}}{1 - \frac{(a + b + c) (a b + b c + c a)}{a b c}} \\ \Rightarrow θ = \tan^{- 1} 0 \Rightarrow \tan θ = 0 \end{aligned}$

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Inverse Circular Functions 3 Question 11

11. If a,b,c are positive real numbers θ=tan−1⁡a(a+b+c)bc+tan−1⁡b(a+b+c)ca

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11. If $a, b, c$ are positive real numbers $θ = \tan^{- 1} \sqrt{\frac{a (a + b + c)}{b c}} + \tan^{- 1} \sqrt{\frac{b (a + b + c)}{c a}}$