Question

If cosecc $\theta$ = $\sqrt{10}$ then sec $\theta$ = ?

(a) $\frac{3}{\sqrt{10}}$ (b) $\frac{\sqrt{10}}{3}$ (c) $\frac{1}{\sqrt{10}}$ (d) $\frac{2}{\sqrt{10}}$

Answer 1

$△$ABC is a right-angled triangle.

Given: $cosec\theta =\sqrt{10}$

But $sin\theta =\frac{1}{cosec\theta }=\frac{1}{\sqrt{10}}$

Also, $sin\theta =\frac{Oppositeside}{Hypotenuseside}=\frac{BC}{AC}\frac{BC}{AC}=\frac{1}{\sqrt{10}}$

Thus $BC=kandAC=\sqrt{10}k$

Using Pythagoras theorem in $△$ ABC, we have,

$A{C}^2=A{B}^2+B{C}^2\Rightarrow A{B}^2=A{C}^2-B{C}^2\Rightarrow A{B}^2=(\sqrt{10}k{)}^2-(k{)}^2\Rightarrow A{B}^2=9k^2\Rightarrow AB=3ksec\theta =\frac{AC}{AB}=\frac{\sqrt{10}k}{3k}=\frac{\sqrt{10}}{3}$