Question

If cosec θ = 2, show that $\left(\mathrm{cot\theta }+\frac{\mathrm{sin\theta }}{1+\mathrm{cos\theta }}\right)=2.$

Answer 1

Let us consider a right $△$ABC, right angled at B and $\angle C=\theta$.
Now, it is given that cosec $\theta$ = 2.
Also, sin $\theta$ = $\frac{1}{\cos ec\theta }=\frac{1}{2}=\frac{AB}{AC}$


So, if AB = k, then AC = 2k, where k is a positive number.
Using Pythagoras theorem, we have:
⇒ AC² = AB² + BC²
⇒ BC² = AC² $-$ AB²
⇒ BC² = (2k)² $-$ (k)²
⇒ BC² = 3k²
⇒ BC = $\sqrt{3}$k
Finding out the other T-ratios using their definitions, we get:
cos $\theta$ = $\frac{BC}{AC}=\frac{\sqrt{3}k}{2k}=\frac{\sqrt{3}}{2}$
tan $\theta$ = $\frac{AB}{BC}=\frac{k}{\sqrt{3}k}=\frac{1}{\sqrt{3}}$
cot $\theta$ = $\frac{1}{\tan \theta }=\sqrt{3}$
Substituting these values in the given expression, we get:
$$\begin{gathered} cot\theta +\frac{\sin \theta }{1+\cos \theta } \\ =\sqrt{3}+\frac{\left({\displaystyle \frac{1}{2}}\right)}{1+{\displaystyle \frac{\sqrt{3}}{2}}} \\ =\sqrt{3}+\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{2+\sqrt{3}}{2}}} \end{gathered}$$

$$\begin{gathered} =\sqrt{3}+\frac{1}{2+\sqrt{3}} \\ =\frac{\sqrt{3}\left(2+\sqrt{3}\right)+1}{2+\sqrt{3}} \\ =\frac{2\sqrt{3}+3+1}{2+\sqrt{3}} \\ =\frac{2\left(2+\sqrt{3}\right)}{2+\sqrt{3}}=2 \end{gathered}$$
i.e., LHS = RHS

Hence proved.