Question

If tanA and tanB are the roots of $x^2-2x-1=0$ , then $si{n}^2(A+B)$ is

• A. $1$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{2}$
• D. $0$

Answer 1

The correct option is A $1$

Given,

$x^2-2x-1=0$

$x=\frac{-(-2)\pm \sqrt{{(-2)}^2-4\cdot 1(-1)}}{2\cdot 1}$

$x=1+\sqrt{2},x=1-\sqrt{2}$

$\Rightarrow \tan A=1+\sqrt{2},\tan B=1-\sqrt{2}$

$\Rightarrow A={\tan }^{-1}(1+\sqrt{2}),B={\tan }^{-1}(1-\sqrt{2})$

$\Rightarrow A=67{\frac{1}{2}}^{\circ },B=22{\frac{1}{2}}^{\circ }$

Now,

${\sin }^2(A+B)$

$={\sin }^2(67{\frac{1}{2}}^{\circ }+22{\frac{1}{2}}^{\circ })$

$={\sin }^2\left({90}^{\circ }\right)$

$=1$