Question

If $e^{({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+.....\infty )\ln 3}$, $xϵ(0,\frac{\pi }{2})$ satisfies the equation $t^2-28t+27=0$ then value of $(\cos x+\sin x{)}^{-1}$ equals

• A. $\sqrt{3}-1$
• B. $2(\sqrt{3}-1)$
• C. $\sqrt{3}+1$
• D. $1$

Answer 1

The correct option is A $\sqrt{3}-1$

\sin ce ${\sin }^{2}x<1$ for $xϵ(0,\pi /2)$
So, $e^{({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+.......\infty )\ln 3}$ $=e^{\left(\frac{{\sin }^{2}x}{1-{\sin }^{2}x}\right)\ln 3}$
$=e^{\left({\tan }^{2}x\right)\ln 3}$ $=e^{\ln 3^{{\tan }^{2}x}}$ $=3^{{\tan }^{2}x}$
Now roots of given quadratic equation is 1 and 27 and also $xϵ(0,\pi /2)$, So
$3^{{\tan }^{2}x}=27$ $\Rightarrow \tan x=\sqrt{3}$
$\Rightarrow \sin x=\frac{\sqrt{3}}{2}$ and $\cos x=\frac{1}{2}$
So, $(\sin x+\cos x{)}^{-1}=\frac{2}{\sqrt{3}+1}$ $=\frac{2}{\sqrt{3}+1}.\frac{\sqrt{3}-1}{\sqrt{3}-1}$ $=\sqrt{3}-1$
Hence, option 'A' is correct.