If $\sin \theta=\frac{\sqrt{3}}{2}$, find the value of all T-ratios of $\theta$.
Let us first draw a right $\triangle \mathrm{ABC}$, right angled at $\mathrm{B}$ and $\angle C=\theta$.
Now, we know that $\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{\sqrt{3}}{2}$.
So, if $\mathrm{AB}=\sqrt{3} k$, then $\mathrm{AC}=2 k$, where $k$ is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2
$\Rightarrow \mathrm{BC}^{2}=\mathrm{AC}^{2}-\mathrm{AB}^{2}=(2 k)^{2}-(\sqrt{3} k)^{2}$
⇒ BC2 = 4k2
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
$\cos \theta=\frac{B C}{A C}=\frac{k}{2 k}=\frac{1}{2}$
$\tan \theta=\frac{A B}{B C}=\frac{\sqrt{3} k}{k}=\sqrt{3}$
$\therefore \cot \theta=\frac{1}{\tan \theta}=\frac{1}{\sqrt{3}}, \operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{2}{\sqrt{3}}$ and $\sec \theta=\frac{1}{\cos \theta}=2$
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