Question

If $\sqrt{3}\tan \mathrm{\theta }-1=0$, find the value of ${\sin }^2\mathrm{\theta }-{\cos }^2\mathrm{\theta }$.

Answer 1

It is given that,
$$\begin{gathered} \sqrt{3}\tan \theta -1=0 \\ \Rightarrow \sqrt{3}\tan \theta =1 \\ \Rightarrow \tan \theta =\frac{1}{\sqrt{3}} \\ \Rightarrow \tan \theta =\tan 30° \\ \Rightarrow \theta =30° \\ \mathrm{Now}, \\ {\sin }^2\theta -{\cos }^2\theta \\ ={\sin }^{2}30°-{\cos }^{2}30° \\ ={\left(\frac{1}{2}\right)}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2 \\ =\left(\frac{1}{4}\right)-\left(\frac{3}{4}\right) \\ =-\frac{1}{2} \end{gathered}$$