Question

STATEMENT 1 : There is no triangle $ABC$ for $A=ta{n}^{-1}2,B=ta{n}^{-1}3$.

STATEMENT 2: lf $x>0,y>0$ and $xy>1$ then $ta{n}^{-1}x+ta{n}^{-1}y=\pi +ta{n}^{-1}\left(\frac{x+y}{1-xy}\right)$

• A. Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of Statement- 1
• B. Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation for Statement-1
• C. Statement-1 is true, Statement-2 is false
• D. Statement-1 is false, Statement-2 is true

Answer 1

Answer: (D)

Statement-1 is false, Statement-2 is true

For a triangle $ABC$, $A+B+C=\pi$
$\Rightarrow A+B=\pi -C$
$\Rightarrow {\tan }^{-1}2+{\tan }^{-1}3=\pi -C$
$\Rightarrow \pi +{\tan }^{-1}\left(\frac{2+3}{1-6}\right)=\pi -C$ (using property of inverse tan function)
$\Rightarrow C=-{\tan }^{-1}(-1)=\frac{\pi }{4}$
Hence a triangle with $A={\tan }^{-1}2,B={\tan }^{-1}3$ is possible. Statement I is false.
Statement II is property of addition of inverse tan function when $x>0,y>0,xy>1$. Hence correct.