If $\cos \theta =\frac{\mathrm{\surd 3}}{2}$, then ${\tan }^2\theta {\cos }^2\theta =?$

The correct answer is $\frac{1}{4}$.

We are given:

$\cos \theta =\frac{\sqrt{3}}{2}$

and we need to find the value of ${\tan }^2\theta ·{\cos }^2\theta$.

Step 1: Find sin θ using the Pythagorean identity.

We know that ${\sin }^2\theta +{\cos }^2\theta =1$, so:

${\sin }^2\theta =1-{\cos }^2\theta =1-{\left(\frac{\sqrt{3}}{2}\right)}^2=1-\frac{3}{4}=\frac{1}{4}$

Therefore, $\sin \theta =\frac{1}{2}$.

Step 2: Find tan θ.

Using the definition $\tan \theta =\frac{\sin \theta }{\cos \theta }$:

$\tan \theta =\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}$

Step 3: Compute tan²θ and cos²θ separately.

${\tan }^2\theta ={\left(\frac{1}{\sqrt{3}}\right)}^2=\frac{1}{3}$

${\cos }^2\theta ={\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3}{4}$

Step 4: Multiply the two results.

${\tan }^2\theta ·{\cos }^2\theta =\frac{1}{3}\times \frac{3}{4}=\frac{3}{12}=\frac{1}{4}$

Alternative (Shortcut) Approach:

Notice that ${\tan }^2\theta ·{\cos }^2\theta$ can be rewritten using the identity $\tan \theta =\frac{\sin \theta }{\cos \theta }$:

${\tan }^2\theta ·{\cos }^2\theta =\frac{{\sin }^2}{\theta }·{\cos }^2\theta ={\sin }^2\theta$

The cos²θ terms cancel, leaving simply ${\sin }^2\theta$. Since $\sin \theta =\frac{1}{2}$, we get:

${\sin }^2\theta ={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$

Therefore, ${\tan }^2\theta ·{\cos }^2\theta =\frac{1}{4}$.