MCQ
Choose the correct answer from the given four options. The value of the expression $\Big[\frac{\sin^222^\circ+\sin^268^\circ}{\cos^222^\circ+\cos^268^\circ}+\sin^263^\circ+\cos63^\circ\sin27^\circ\Big]$ is :
- A$3$
- ✓$2$
- C$1$
- D$0$
✓
Answer
Correct option: B.
$2$
Given expression, $\frac{\sin^222^\circ+\sin^268^\circ}{\cos^222^\circ+\cos^268^\circ}+\sin^263^\circ+\cos63^\circ\sin27^\circ$
$=\frac{\sin^222^\circ+\sin^2(90^\circ-22^\circ)}{\cos^2(90^\circ-68^\circ)+\cos^268^\circ}+\sin^263+\cos63^\circ\sin(90^\circ-63^\circ)$
$=\frac{\sin^222^\circ+\cos^222^\circ}{\sin^268^\circ+\cos^268^\circ}+\sin^263^\circ+\cos63^\circ.\cos63^\circ$
$\begin{bmatrix}\because\ \sin(90^\circ-\theta)=\cos\theta \\\text{and }\cos(90^\circ-\theta)=\sin\theta \end{bmatrix}$
$=\frac{1}{1}+(\sin^263^\circ+\cos^263^\circ)$ $[\because\sin^2\theta+\cos^2\theta=1]$
$=1+1=2$
$=\frac{\sin^222^\circ+\sin^2(90^\circ-22^\circ)}{\cos^2(90^\circ-68^\circ)+\cos^268^\circ}+\sin^263+\cos63^\circ\sin(90^\circ-63^\circ)$
$=\frac{\sin^222^\circ+\cos^222^\circ}{\sin^268^\circ+\cos^268^\circ}+\sin^263^\circ+\cos63^\circ.\cos63^\circ$
$\begin{bmatrix}\because\ \sin(90^\circ-\theta)=\cos\theta \\\text{and }\cos(90^\circ-\theta)=\sin\theta \end{bmatrix}$
$=\frac{1}{1}+(\sin^263^\circ+\cos^263^\circ)$ $[\because\sin^2\theta+\cos^2\theta=1]$
$=1+1=2$
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