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 that $\tan {45}^0=1$ ,$\cos {30}^0=\frac{\sqrt{3}}{2}$

or $\cos {45}^0=\frac{1}{\sqrt{2}}$ ,$\sin {45}^0=\frac{1}{\sqrt{2}}$

Then $1^2-(\frac{\sqrt{3}}{2}{)}^2=x\cdot \frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}$

$\Rightarrow 1-\frac{3}{4}=x\times (\frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}})\Rightarrow \frac{x}{2}=\frac{1}{4}\Rightarrow x=\frac{1}{2}$