Question

Determine whether the following equation is identity or not:

$\frac{\mathrm{tan\theta }+\mathrm{sin\theta }}{\mathrm{tan\theta }-\mathrm{sin\theta }}=\frac{\mathrm{sec\theta }+1}{\mathrm{sec\theta }-1}$.

Answer 1

Verify L.H.S and R.H.S at any value of $\mathit{\theta }$

$\frac{\mathrm{tan\theta }+\mathrm{sin\theta }}{\mathrm{tan\theta }-\mathrm{sin\theta }}=\frac{\mathrm{sec\theta }+1}{\mathrm{sec\theta }-1}$

Let, $\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{tan\theta }+\mathrm{sin\theta }}{\mathrm{tan\theta }-\mathrm{sin\theta }}$

When $\mathrm{\theta }={60}^{\circ }$

$\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\tan {60}^{\circ }+\sin {60}^{\circ }}{\tan {60}^{\circ }-\sin {60}^{\circ }}$

$$\begin{gathered} =\frac{\sqrt{3}+{\displaystyle \frac{\sqrt{3}}{2}}}{\sqrt{3}-{\displaystyle \frac{\sqrt{3}}{2}}} \\ =\frac{\sqrt{3}\left(1+{\displaystyle \frac{1}{2}}\right)}{\sqrt{3}\left(1-\frac{1}{2}\right)} \\ =\frac{\sqrt{3}\times {\displaystyle \frac{3}{2}}}{\sqrt{3}\times {\displaystyle \frac{1}{2}}} \\ =3 \end{gathered}$$

$\mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{\sec {60}^{\circ }+1}{\sec {60}^{\circ }-1}$

$$\begin{gathered} =\frac{2+1}{2-1} \\ =\frac{3}{1} \end{gathered}$$

$\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.$

Hence, the given equation is identity.