Question

If $4\sin \theta =3$ and $\sqrt{\frac{{\csc }^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}}+23\cot \theta =\frac{\sqrt{7}}{x}+\cos \theta$, then find the value of $x$

Answer 1

$\begin{array}{c}4\sin \theta =3\\ \sin \theta =\frac{3}{4}\\ So,\cos \theta =\frac{\sqrt{7}}{4}\\ Now,\\ \sqrt{\frac{\cos e{c}^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}}+23\cot \theta =\frac{\sqrt{7}}{x}+\cos \theta \\ \sqrt{\frac{1}{{\tan }^2\theta }}+23\cot \theta =\frac{\sqrt{7}}{x}+\cos \theta \\ \cot \theta +23\cot \theta =\frac{\sqrt{7}}{x}+\cos \theta \\ \frac{24\sqrt{7}}{3}=\frac{\sqrt{7}}{x}+\frac{\sqrt{7}}{4}\\ 8\sqrt{7}=\sqrt{7}\left(\frac{1}{x}+\frac{1}{4}\right)\\ \frac{1}{x}+\frac{1}{4}=8\\ \frac{1}{x}=8-\frac{1}{4}\\ \frac{1}{x}=\frac{31}{4}\\ x=\frac{4}{31}\end{array}$

Hence, this is the answer.