Question

If $\csc \theta +\cot \theta =c$, then what is the $\cos \theta =$

• A. $\frac{c}{c^2-1}$
• B. $\frac{c}{c^2+1}$
• C. $\frac{c^2-1}{c^2+1}$
• D. None of the above

Answer 1

The correct option is C $\frac{c^2-1}{c^2+1}$

$\csc \theta +\cot \theta =c$

$\Rightarrow \frac{1+\cos \theta }{\sin \theta }=c$

$\Rightarrow 1+2\cos \theta +{\cos }^2\theta =c^2{\sin }^2\theta$

$\Rightarrow (1-c^2)+2\cos \theta +(1+c^2){\cos }^2\theta =0$

We have a formula for solving quadratic equation $a{x}^2+bx+c=0$ is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow \cos \theta =\frac{-2\pm \sqrt{4-4(1-c^2)(1+c^2)}}{2(1+c^2)}$

$\Rightarrow \cos \theta =\frac{-2\pm 2c^2}{2(1+c^2)}$

$\therefore \cos \theta =\frac{c^2-1}{c^2+1}$ or $\cos \theta =-1$