Question

\(\tan^{-1}\!\left(\dfrac{x+y}{1-xy}\right)=\ \ ?\)

• A. \(\sin^{-1}(x+y)\)
• B. \(\cos^{-1}(x+y)\)
• C. \(\tan^{-1}(x+y)\)
• D. \(\tan^{-1}x+\tan^{-1}y\)

Hint:

Spot $\dfrac{x+y}{1-xy}$ $\Rightarrow$ it is $\tan(\alpha+\beta)$ in disguise.

Answer 1

The Correct Option is D

Use the tangent addition formula \(\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\,\tan\beta}\). Take \(\alpha=\tan^{-1}x,\ \beta=\tan^{-1}y\). Then \(\tan(\alpha+\beta)=\dfrac{x+y}{1-xy}\). Apply \(\tan^{-1}\): \(\alpha+\beta=\tan^{-1}x+\tan^{-1}y\).