if cotθ=3x-1/12x , then show that cotθ+cosecθ=6x or -1/6x

Question

parthkohli · Answer

First of all, welcome back to OS.

Secondly, find the value of $\csc \theta $ by using the relation $\csc ^2 \theta = 1 + \cot^2 \theta$.

anonymous · Answer

cot (x) is the contangent function.

anonymous · Answer

@dumbcow

mathmath333 · Answer

$\large\tt \color{black}{cot\theta =3x-\dfrac{1}{12x}-->A}$

$\large\tt \color{black}{(cot^{2}\theta) =(3x-\dfrac{1}{12x})^2}$


$\large\tt \color{black}{(cot^{2} \theta) =(9x^{2}-\dfrac{2\times3x}{12x}+\dfrac{1}{144x^2})}$


$\large\tt \color{black}{cot^{2}\theta =(9x^2-\dfrac{1}{2}+\dfrac{1}{144x^2})}$


$\large\tt \color{black}{cosec^{2}\theta-1 =(9x^2-\dfrac{1}{2}+\dfrac{1}{144x^2})}$


$\large\tt \color{black}{cosec^{2}\theta =(9x^2-\dfrac{1}{2}+\dfrac{1}{144x^2})+1}$



$\large\tt \color{black}{cosec^{2}\theta =(9x^2+\dfrac{1}{2}+\dfrac{1}{144x^2})}$


$\large\tt \color{black}{cosec^{2}\theta =(3x+\dfrac{1}{12x})^2}$



$\large\tt \color{black}{cosec\theta =\pm(3x+\dfrac{1}{12})-->B}$


$\large\tt \color{black}{from~~A+B=cosec\theta+cot\theta =+(3x+\dfrac{1}{12x})+(3x-\dfrac{1}{12x})=6x}$



$\large\tt \color{black}{and~~A+B=cosec\theta+cot\theta =-(3x+\dfrac{1}{12x})+(3x-\dfrac{1}{12x})=\dfrac{-2}{12x}=\dfrac{-1}{6x}}$

parthkohli · Answer

Yeah...

mathmath333 · Answer

$\Huge\tt \color{black}{\ddot\smile}$