Inverse Circular Functions 3 Question 12
12. Solve the following equation for $x$.
$$ \tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4} $$
(1978, 3M)
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Answer:
Correct Answer: 12. $x=\frac{1}{6}$
Solution:
- Given, $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$
$$ \begin{aligned} & \Rightarrow \quad \tan ^{-1} \frac{2 x+3 x}{1-6 x^{2}}=\frac{\pi}{4} \\ & \Rightarrow \quad \frac{5 x}{1-6 x^{2}}=1 \end{aligned} $$
$$ \Rightarrow \quad 6 x^{2}+5 x-1=0 $$
$$ \Rightarrow \quad(x+1)(6 x-1)=0 $$
$\Rightarrow$
$$ x=-1 \text { or } \frac{1}{6} $$
But $x=-1$ does not satisfy the given equation.
$\therefore$ We take $x=\frac{1}{6}$
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