Question

The solution of the trigonometric equation
$\frac{{\cos }^2\theta }{{\cot }^2\theta -{\cos }^2\theta }=3,0^{\circ }<\theta <{90}^{\circ }$

• A. $\theta =0^{\circ }$
• B. $\theta ={30}^{\circ }$
• C. $\theta ={60}^{\circ }$
• D. $\theta ={90}^{\circ }$

Answer 1

Answer: (C)

$\theta ={60}^{\circ }$

Simplifying, we get
$co{s}^2\theta =3co{t}^2\theta -3co{s}^2\theta$
$4co{s}^2\theta -3co{t}^2\theta =0$
$co{s}^2\theta (4-\frac{3}{si{n}^2\theta })=0$
Hence
$co{s}^2\theta =0$
$\theta ={90}^0$
And
$4-\frac{3}{si{n}^2\theta }=0$
$si{n}^2\theta =\frac{3}{4}$
$sin\theta =\pm \frac{\sqrt{3}}{2}$
Hence
$\theta =\pm {60}^0$
Now
$0<\theta <{90}^0$
Hence
$\theta ={60}^0$