Question

Determine whether the following equation is identity or not:

${\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }=\frac{2\mathrm{tan\theta }}{1+{\tan }^2\mathrm{\theta }}$.

Answer 1

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

${\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }=\frac{2\mathrm{tan\theta }}{1+{\tan }^2\mathrm{\theta }}$

Let, $\mathrm{L}.\mathrm{H}.\mathrm{S}.={\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }$

When, $\mathrm{\theta }={30}^{\circ }$

$\mathrm{L}.\mathrm{H}.\mathrm{S}.={\cos }^2{30}^{\circ }-{\sin }^2{30}^{\circ }$

$$\begin{gathered} ={\left(\frac{\sqrt{3}}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2 \\ =\frac{3}{4}-\frac{1}{4} \\ =\frac{2}{4} \\ =\frac{1}{2} \end{gathered}$$

$\mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{2\tan {30}^{\circ }}{1+{\tan }^2{30}^{\circ }}$

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

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

Hence, the given equation is not an identity.