# If tan2 45° − cos2 30° = x sin 45° cos 45°, then x =

Question:

If $\tan ^{2} 45^{\circ}-\cos ^{2} 30^{\circ}=x \sin 45^{\circ} \cos 45^{\circ}$, then $x=$

(a) 2

(b) $-2$

(c) $-\frac{1}{2}$

(d) $\frac{1}{2}$

Solution:

We are given: $\tan ^{2} 45^{\circ}-\cos ^{2} 30^{\circ}=x \sin 45^{\circ} \cos 45^{\circ}$

We have to find $x$

$\Rightarrow 1-\left(\frac{\sqrt{3}}{2}\right)^{2}=x \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$

$\Rightarrow 1-\frac{3}{4}=\frac{x}{2}$

$\Rightarrow \frac{1}{4}=\frac{x}{2}$

$\Rightarrow x=\frac{1}{2}$

We know that $\left[\begin{array}{l}\sin 45^{\circ}=\frac{1}{\sqrt{2}} \\ \cos 45^{\circ}=\frac{1}{\sqrt{2}} \\ \tan 45^{\circ}=1 \\ \cos 30^{\circ}=\frac{\sqrt{3}}{2}\end{array}\right]$

Hence the correct option is $(d)$