MCQ
If $\tan^245^\circ-\cos^230^\circ=\text{x}\sin45^\circ\cos45^\circ,$ then $x =$
- A$2$
- B$-2$
- C$-\frac{1}{2}$
- ✓$\frac{1}{2}$
✓
Answer
Correct option: D.
$\frac{1}{2}$
We have,
$\tan^245^\circ-\cos^230^\circ$
$=\text{x}\sin45^\circ\cos45^\circ\dots(1)$
Put the values in $(1)$
$\Rightarrow(1)^2-\Big(\frac{\sqrt{3}}{2}\Big)^2$
$=\text{x}\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}$
$\Rightarrow1-\frac{3}{4}=\text{x}\times\frac{1}{2}$
$\Rightarrow\frac{1}{4}=\frac{\text{x}}{2}$
$\Rightarrow\text{x}=\frac{1}{2}$
Thus, $\text{x}=\frac{1}{2}$
Hence the correct option is $(d)$
$\tan^245^\circ-\cos^230^\circ$
$=\text{x}\sin45^\circ\cos45^\circ\dots(1)$
Put the values in $(1)$
$\Rightarrow(1)^2-\Big(\frac{\sqrt{3}}{2}\Big)^2$
$=\text{x}\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}$
$\Rightarrow1-\frac{3}{4}=\text{x}\times\frac{1}{2}$
$\Rightarrow\frac{1}{4}=\frac{\text{x}}{2}$
$\Rightarrow\text{x}=\frac{1}{2}$
Thus, $\text{x}=\frac{1}{2}$
Hence the correct option is $(d)$
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