If \(\sqrt{3}\,\tan\theta=1\), find \(\sin^2\theta-\cos^2\theta\).
\(\displaystyle \sin^2\theta-\cos^2\theta=-\dfrac{1}{2}.\)
Step 1: We are given: \(\sqrt{3}\,\tan\theta = 1\).
Divide both sides by \(\sqrt{3}\):
\(\tan\theta = \dfrac{1}{\sqrt{3}}\).
Step 2: Recall the standard values of tan:
So, \(\theta = 30^\circ\) (since we consider an acute angle).
Step 3: Now we need to calculate \(\sin^2\theta - \cos^2\theta\).
Substitute \(\theta = 30^\circ\):
\(\sin^2 30^\circ - \cos^2 30^\circ\).
Step 4: Use the standard values of sine and cosine:
Step 5: Square them:
\(\sin^2 30^\circ = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}\)
\(\cos^2 30^\circ = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3}{4}\)
Step 6: Subtract:
\(\sin^2 30^\circ - \cos^2 30^\circ = \dfrac{1}{4} - \dfrac{3}{4} = -\dfrac{2}{4} = -\dfrac{1}{2}.\)
Final Answer: \(\sin^2\theta - \cos^2\theta = -\dfrac{1}{2}\).