Question

If $\cot \theta =3x-\frac{1}{12x}$ then show that $\csc \theta +\cot \theta =6x$ or $-\frac{1}{6x}$.

Answer 1

$\cot \theta =3x-\frac{1}{12x}$

Squaring both sides, we have

${\cot }^2\theta ={(3x-\frac{1}{12x})}^2$

${\csc }^2\theta -1=9x^2+\frac{1}{144x^2}-\frac{1}{2}$

$\Rightarrow {\csc }^2\theta =9x^2+\frac{1}{144x^2}-\frac{1}{2}+1$

$\Rightarrow {\csc }^2\theta =9x^2+\frac{1}{144x^2}+\frac{1}{2}$

$\Rightarrow {\csc }^2\theta ={(3x+\frac{1}{12x})}^2$

$\Rightarrow \csc \theta =\pm (3x+\frac{1}{12x})$

Now,

Case I:- $\csc \theta =3x+\frac{1}{12x}$

Therefore,

$\cot \theta +\csc \theta =(3x-\frac{1}{12x})+(3x+\frac{1}{12x})=6x$

Case II:- $\csc \theta =-(3x+\frac{1}{12x})$

Therefore,

$\cot \theta +\csc \theta =(3x-\frac{1}{12x})+(-3x-\frac{1}{12x})=-\frac{1}{6x}$

Hence proved.