Question

If ${\log }_{\cos x}\tan x+{\log }_{\sin x}\cot x=0$ , then the most general solution of $x$ is

• A. $n\pi +\frac{\pi }{4},n\in Z$
• B. $2n\pi +\frac{\pi }{4},n\in Z$
• C. $2n\pi +\frac{3\pi }{4},n\in Z$
• D. none of these

Answer 1

Answer: (B)

$2n\pi +\frac{\pi }{4},n\in Z$

$n\in Z$
For ${\log }_a\left(x\right);x,a>0$ but $a\ne 1$
Using this, we can say that $\sin \left(x\right),\cos \left(x\right)>0$, which gives us $x$ in range $2n\pi +(0,\frac{\pi }{2})$ i.e. first Quadrant....... {1}
But $\tan \left(x\right),\cot \left(x\right)$ are always positive in first quadrant, so for the expression ${\log }_{\cos \left(x\right)}\tan \left(x\right)+{\log }_{\sin \left(x\right)}\cot \left(x\right)=0,[\because \log 1=0]$ to be valid, both the terms has to be zero, implies $\tan \left(x\right),\cot \left(x\right)=1$, which gives $x=2n\pi +\frac{\pi }{4}$ .... {2}
Taking intersection of {1} and {2},
$x=2n\pi +\frac{\pi }{4},n\in Z$
Hence, option B.