Question

If cosec θ = $\sqrt{10}$, the find the values of all T-ratios of θ.

Answer 1

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that cosec $\theta$ = $\frac{\mathrm{Hypotenuse}}{\mathrm{Perpendicular}}$ = $\frac{AC}{AB}$= $\frac{\sqrt{10}}{1}$.

So, if AC = ($\sqrt{10}$)k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ BC² = AC² $-$ AB² = 10k² $-$ k²
⇒ BC² = 9k²
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
tan $\theta$ = $\frac{AB}{BC}$ = $\frac{k}{3k}=\frac{1}{3}$

cos $\theta$ = $\frac{BC}{AC}=\frac{3k}{\sqrt{10}k}=\frac{3}{\sqrt{10}}$

∴ $\sin \theta =\frac{1}{\cos ec\theta }=\frac{1}{\sqrt{10}}$, cot $\theta$ = $\frac{1}{\tan \theta }=3$ and sec $\theta$ = $\frac{1}{\cos \theta }=\frac{\sqrt{10}}{3}$