Question

The smallest positive value of $x$ which satisfies the equation ${\log }_{\cos x}\sin x+{\log }_{\sin x}\cos x=2$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is C $\frac{\pi }{4}$

${\log }_{\cos x}\sin x+{\log }_{\sin x}\cos x=2$
Let ${\log }_{\cos x}\sin x=t$
$\Rightarrow t+\frac{1}{t}=2\Rightarrow t^2-2t+1=0$
$\Rightarrow (t-1{)}^2=0\Rightarrow t=1$
$\Rightarrow {\log }_{\cos x}\sin x=1\text{or}\sin x=\cos x\Rightarrow \tan x=1$
$\Rightarrow x=\frac{\pi }{4}$