Question

If sin $\mathrm{\theta }=\frac{\sqrt{3}}{2}$, find the value 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 sin $\theta$ = $\frac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}$= $\frac{AB}{AC}$ = $\frac{\sqrt{3}}{2}$.

So, if AB = $\sqrt{3}k$, then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ BC² = AC² $-$ AB² = (2k)² $-$ ($\sqrt{3}k$)²
⇒ BC² = 4k² $-$ 3k² = k²
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
cos $\theta$ = $\frac{BC}{AC}$ = $\frac{k}{2k}=\frac{1}{2}$
tan $\theta$ = $\frac{AB}{BC}=\frac{\sqrt{3}k}{k}=\sqrt{3}$

∴ cot $\theta$ = $\frac{1}{\tan \theta }=\frac{1}{\sqrt{3}}$, cosec $\theta$ = $\frac{1}{\sin \theta }=\frac{2}{\sqrt{3}}$ and sec $\theta$ = $\frac{1}{\cos \theta }=2$