SATHEE: Inverse Circular Functions 3 Question 12

Inverse Circular Functions 3 Question 12

12. Solve the following equation for $x$ .

$\tan^{- 1} 2 x + \tan^{- 1} 3 x = \frac{π}{4}$

(1978, 3M)

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Answer:

Correct Answer: 12. $x = \frac{1}{6}$

Solution:

Given, $\tan^{- 1} 2 x + \tan^{- 1} 3 x = \frac{π}{4}$

$\begin{aligned} \Rightarrow \tan^{- 1} \frac{2 x + 3 x}{1 - 6 x^{2}} = \frac{π}{4} \\ \Rightarrow \frac{5 x}{1 - 6 x^{2}} = 1 \end{aligned}$

$\Rightarrow 6 x^{2} + 5 x - 1 = 0$

$\Rightarrow (x + 1) (6 x - 1) = 0$

$\Rightarrow$

$x = - 1 or \frac{1}{6}$

But $x = - 1$ does not satisfy the given equation.

$∴$ We take $x = \frac{1}{6}$

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Inverse Circular Functions 3 Question 12

12. Solve the following equation for x.

Answer:

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12. Solve the following equation for $x$ .