Question

If sin$\theta -cosec\theta =\sqrt{5},$ then the value of sin $\theta +cosec\theta$ is:

• A. $\sqrt{3}$
• B. 1
• C. 3
• D. 9

Answer 1

Answer: (C)

3

$\Rightarrow \sin \theta -cosec\theta =\sqrt{5}$

$\Rightarrow \sin \theta -\frac{1}{\sin \theta }=\sqrt{5}$ $(\because cosec\theta =\frac{1}{\sin \theta })$

$\Rightarrow {\sin }^2\theta -\sqrt{5}\sin \theta -1=0$

Solving equation to get roots.

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{\pm 3+\sqrt{5}}{2}$ (substitute values to get roots)

To find:-

$\sin \theta +cosec\theta$

Using $\frac{3+\sqrt{5}}{2}$

$=\frac{3+\sqrt{5}}{2}+\frac{2}{3+\sqrt{5}}$

$=\frac{(3+\sqrt{5}{)}^2+4}{2(3+\sqrt{5})}$

$=\frac{9+4+5+6\sqrt{5}}{2(3+\sqrt{5})}$

$=\frac{6(3+\sqrt{5})}{2(3+\sqrt{5})}$

$=3$

Using $\frac{-3+\sqrt{5}}{2}$

$=\frac{-3+\sqrt{5}}{2}+\frac{2}{-3+\sqrt{5}}$

$=\frac{(-3+\sqrt{5}{)}^2+4}{2(-3+\sqrt{5})}$

$=\frac{9+4+5-6\sqrt{5}}{2(-3+\sqrt{5})}$

$=\frac{-6(-3+\sqrt{5})}{2(-3+\sqrt{5})}$

$=-3$

According to option answer is $3$