Question

Determine whether the following equation is identity or not:

${\cot }^2\mathrm{\theta }+\mathrm{cos\theta }={\sin }^2\mathrm{\theta }$.

Answer 1

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

${\cot }^2\mathrm{\theta }+\mathrm{cos\theta }={\sin }^2\mathrm{\theta }$

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

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

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

$$\begin{gathered} ={\cot }^2{30}^{\circ }+\cot {30}^{\circ } \\ ={\left(\sqrt{3}\right)}^2+\frac{\sqrt{3}}{2} \\ =3+\frac{\sqrt{3}}{2} \\ =\frac{6+\sqrt{3}}{2} \end{gathered}$$

$\mathrm{L}.\mathrm{H}.\mathrm{S}.=\left(\sqrt{3}+\frac{1}{2}\right)$

$\mathrm{R}.\mathrm{H}.\mathrm{S}.={\sin }^2\mathrm{\theta }$

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

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

Hence, the given equation is not identity.