Question

The value of ${\cos }^{-1}\left\{\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\surd (1-\sin x)-\surd (1+\sin x)}\right\}$ is $(0<x<2\pi )$

• A. $\pi -\frac{x}{2}$
• B. $2\pi -x$
• C. $-\frac{x}{2}$
• D. $2\pi -\frac{X}{2}$

Answer 1

The correct option is C $-\frac{x}{2}$

$co{t}^{-1}\left(\frac{\sqrt{1-sinx}+\sqrt{1+sinx}}{\sqrt{1-sinx}-\sqrt{1+sinx}}\right)$

$sin2x=2sinxcosx.$

also, $sinx=2sin\frac{x}{2}cos\frac{x}{2}...\left(1\right)$

also, $1+sinx=2sin\frac{x}{2}cos\frac{x}{2}+1$

since $si{n}^{2}x+co{s}^{2}x=1$

$si{n}^2\frac{x}{2}+co{s}^2\frac{x}{2}=1$

$1+sinx=si{n}^2\frac{x}{2}+co{s}^2\frac{x}{2}+2sin\frac{x}{2}cos\frac{x}{2}$

$1+sinx=(sin\frac{x}{2}+cos\frac{x}{2}{)}^2$

$\sqrt{1+sinx}=sin\frac{x}{2}+cos\frac{x}{2}...\left(2\right)$

Similarly

multiply by -1 in $e{q}^n$ (1),

$-sinx=2sin\frac{x}{2}cos\frac{x}{2}$

Now adding 1 to the above $e{q}^n$

$1-sinx=1-2sin\frac{x}{2}cos\frac{x}{2}$

$\therefore$ we get $\sqrt{1-sinx}=(cos\frac{x}{2}-sin\frac{x}{2})...\left(3\right)$

$co{s}^{-1}\left(\frac{\sqrt{1-sinx}+\sqrt{1+sinx}}{\sqrt{1-sinx}-\sqrt{1+sinx}}\right)$

$=co{s}^{-1}\left(\frac{cos\frac{x}{2}-sin\frac{x}{2}+sin\frac{x}{2}+cos\frac{x}{2}}{cos\frac{x}{2}-sin\frac{x}{2}-sin\frac{x}{2}-cos\frac{x}{2}}\right)$

$=co{s}^{-1}\left(\frac{2cos\frac{x}{2}}{-2sin\frac{x}{2}}\right)=co{t}^{-1}\left(cot\frac{x}{2}\right)$

$=-co{t}^{-1}\left(cot\frac{x}{2}\right)(\therefore cotx$ is an add function)

$=-\frac{x}{2}$