Question

If $\sin (90-\theta )+\cos \theta =\sqrt{2}\cos (90-\theta )$, for $0<\theta <{90}^{\circ }$, find the value of $\text{cosec}\theta$.

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\sqrt{\frac{2}{3}}$

Answer 1

Answer: (C)

$\sqrt{\frac{3}{2}}$

$\sin (90-\theta )+\cos \theta =\sqrt{2}\cos (90-\theta )\Rightarrow 2cos\theta =\sqrt{2}\sin \theta \Rightarrow \cot \theta =\frac{1}{\sqrt{2}}$

we know that, ${\text{cosec}}^2\theta =1-{\cot }^2\theta \Rightarrow \text{cosec}\theta =\sqrt{1+{\cot }^2\theta }\Rightarrow \text{cosec}\theta =\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}$

Therefore, Answer is $\sqrt{\frac{3}{2}}$