Question

Let $f\left(x\right)=\frac{1-\sin 2x+\cos 2x}{2\cos 2x}$ and $\alpha +\beta =\frac{5\pi }{4}.$ Then the value of $f\left(\alpha \right)\cdot f\left(\beta \right)$ is

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

Answer 1

The correct option is B $\frac{1}{2}$

$f\left(x\right)=\frac{1-\sin 2x+\cos 2x}{2\cos 2x}$
$\Rightarrow f\left(x\right)=\frac{2{\cos }^{2}x-2\sin x\cos x}{2({\cos }^{2}x-{\sin }^{2}x)}$
$\Rightarrow f\left(x\right)=\frac{\cos x}{\cos x+\sin x}=\frac{1}{1+\tan x}$

Now, $f\left(\alpha \right)\cdot f\left(\beta \right)=\left(\frac{1}{1+\tan \alpha }\right)\left(\frac{1}{1+\tan (\frac{5\pi }{4}-\alpha )}\right)$
$=\left(\frac{1}{1+\tan \alpha }\right)\left(\frac{1}{1+\frac{1-\tan \alpha }{1+\tan \alpha }}\right)$
$=\left(\frac{1}{1+\tan \alpha }\right)\left(\frac{1+\tan \alpha }{2}\right)$
$=\frac{1}{2}$