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. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

We know, $\tan {45}^{\circ }=1,\cos 30=\frac{\sqrt{3}}{2},\sin {45}^{\circ }=\frac{1}{\sqrt{2}},\cos {45}^{\circ }=\frac{1}{\sqrt{2}}$

Now, ${\tan }^2{45}^o-{\cos }^2{30}^o=x\sin {45}^0\cos {45}^o$

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

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