Aspire Faculty ID #15473 · Topic: MAH MCA CET 2025 (Shift 1) · Just now
MAH MCA CET 2025 (Shift 1)

If $\tan^{-1}\left(\dfrac{1-x}{1+x}\right)=\dfrac{1}{2}\tan^{-1}x$, then $x$ is

Solution

Let $t=\tan\left(\dfrac{1}{2}\tan^{-1}x\right)=\dfrac{1-x}{1+x}$. Using $\tan(2\theta)=\dfrac{2t}{1-t^2}=x$, we get $x=\dfrac{1-x^2}{2x}\Rightarrow 3x^2=1\Rightarrow x=\pm\dfrac{1}{\sqrt3}$. From the equation, $x>0$.

Previous 10 Questions — MAH MCA CET 2025 (Shift 1)

Nearest first

Next 10 Questions — MAH MCA CET 2025 (Shift 1)

Ascending by ID
Ask Your Question or Put Your Review.

loading...