MHT CET · Maths · Inverse Trigonometric Functions
If \(\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}\), then value of \(x\) is
- A \(\frac{1}{6}\)
- B \(-\frac{1}{6}\)
- C 1
- D \(\frac{5}{6}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{6}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4} \\ & \Rightarrow \tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)=\tan ^{-1}(1) \\ & \Rightarrow \frac{5 x}{1-6 x^2}=1 \\ & \Rightarrow 6 x^2+5 x-1=0 \\ & \Rightarrow(6 x-1)(x+1)=0 \\ & \Rightarrow x=\frac{1}{6} \text { or } x=-1 \\ & \Rightarrow x=\frac{1}{6} \text { satisfies the equation }\end{aligned}\)
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