KCET · Maths · Inverse Trigonometric Functions
If \(\tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{3}\right)\), then \(x\) is
- A \(\frac{1}{3}\)
- B \(\frac{1}{2}\)
- C \(\frac{1}{4}\)
- D \(\frac{1}{6}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Given equation \(\tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{3}\right)\)
\[
\begin{aligned}
&\Rightarrow \tan ^{-1} x=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{3}\right) \\
&\left.\Rightarrow \quad \tan ^{-1} x=\tan ^{-1}\left(\frac{1-\frac{\pi}{3}}{1+\frac{1}{3}}\right) \tan ^{-1}(1)\right) \\
&\Rightarrow \quad\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right] \\
&\Rightarrow \quad \tan ^{-1} x=\tan ^{-1}\left(\frac{2 / 3}{4 / 3}\right) \\
&\Rightarrow \quad x=1 / 2
\end{aligned}
\]
\[
\begin{aligned}
&\Rightarrow \tan ^{-1} x=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{3}\right) \\
&\left.\Rightarrow \quad \tan ^{-1} x=\tan ^{-1}\left(\frac{1-\frac{\pi}{3}}{1+\frac{1}{3}}\right) \tan ^{-1}(1)\right) \\
&\Rightarrow \quad\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right] \\
&\Rightarrow \quad \tan ^{-1} x=\tan ^{-1}\left(\frac{2 / 3}{4 / 3}\right) \\
&\Rightarrow \quad x=1 / 2
\end{aligned}
\]
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