Question

If ${\sin }^{-1}x=\theta +\beta$ and ${\sin }^{-1}y=\theta -\beta$, then $1+xy=$

• A. $si{n}^2\theta +si{n}^2\beta$
• B. $si{n}^2\theta +co{s}^2\beta$
• C. $co{s}^2\theta +co{s}^2\beta$
• D. $co{s}^2\theta +si{n}^2\beta$

Answer 1

The correct option is B $si{n}^2\theta +co{s}^2\beta$

Given, $si{n}^{-1}x=\theta +\beta andsi{n}^{-1}y=\theta -\beta$

$⟹x=sin(\theta +\beta ),y=sin(\theta -\beta )$

$\therefore 1+xy$

$=1+sin(\theta +\beta )sin(\theta -\beta )$

$\because \sin (A-B)=\sin A\cos B-\sin B\cos A$

$\sin (A+B)=\sin A\cos B+\sin B\cos A$

$\therefore 1+xy=1+\left[\right(sin\theta cos\beta +cos\theta sin\beta \left)\right(sin\theta cos\beta -cos\theta sin\beta \left)\right]$

$=1+\left[\right(sin\theta cos\beta {)}^2-\left(cos\theta sin\beta {)}^2\right]$ ....... using $a^2-b^2=(a-b)(a+b)$

$=si{n}^2\left(\theta \right)+co{s}^2\left(\theta \right)+(sin\theta cos\beta {)}^2-(cos\theta sin\beta {)}^2$ .......... $[sin²x+cos²x=1]$

$=si{n}^2\left(\theta \right)[1+co{s}^2\beta ]+co{s}^2\left(\theta \right)[1-si{n}^2\beta ]$

$=si{n}^2\left(\theta \right)+si{n}^2\left(\theta \right)co{s}^2\left(\beta \right)+co{s}^2\left(\theta \right)\left(co{s}^2\beta \right)$

$=si{n}^2\left(\theta \right)+co{s}^2\left(\beta \right)\left[si{n}^2\right(\theta )+co{s}^2(\theta \left)\right]$

$=si{n}^2\left(\theta \right)+co{s}^2\left(\beta \right)$ .......... $[sin²x+cos²x=1]$