Search any question & find its solution
Question:
Answered & Verified by Expert
If $\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}$, where $x>0$, then $x=$
Options:
Solution:
1119 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{6}$
$$
\begin{aligned}
& \tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4} \\
& \therefore \tan ^{-1}\left[\frac{2 x+3 x}{1-(2 x)(3 x)}\right]=\frac{\pi}{4} \Rightarrow \tan \frac{\pi}{4}=\frac{5 x}{1-6 x^2}=1 \\
& \therefore 6 x^2+5 x-1=0 \Rightarrow(6 x-1)(x+1)=0 \Rightarrow x=-1, \frac{1}{6}
\end{aligned}
$$
Since $x>0$, we get $x=\frac{1}{6}$
\begin{aligned}
& \tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4} \\
& \therefore \tan ^{-1}\left[\frac{2 x+3 x}{1-(2 x)(3 x)}\right]=\frac{\pi}{4} \Rightarrow \tan \frac{\pi}{4}=\frac{5 x}{1-6 x^2}=1 \\
& \therefore 6 x^2+5 x-1=0 \Rightarrow(6 x-1)(x+1)=0 \Rightarrow x=-1, \frac{1}{6}
\end{aligned}
$$
Since $x>0$, we get $x=\frac{1}{6}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.