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 $(d)$