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Model Paper 5 — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSModel Paper 51 Mark
MCQ
The value of $\sin ^2 \frac{5 \pi}{12}-\sin ^2 \frac{\pi}{12}$ is
  • A
    $\sqrt{3} / 2$
  • B
    $\frac{1}{2}$
  • C
    $0$
  • D
    1
✓

Answer

(a) $\sqrt{3} / 2$
Explanation: $\frac{5 \pi}{12}=75^{\circ}, \frac{\pi}{12}=15^{\circ}$
$\begin{array}{l}\sin ^2 75^{\circ}-\sin ^2 15^{\circ} \\ =\sin ^2 75^{\circ}-\cos ^2 75^{\circ}\left[\sin \left(90^{\circ}-\theta\right)=\cos \theta\right]\end{array}$
Now, $\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)$
$\begin{array}{l}=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ} \\ =\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2} \\ =\frac{\sqrt{3}+1}{2 \sqrt{2}}\end{array}$
$\cos 75^{\circ}=\cos \left(45^{\circ}+30^{\circ}\right)$
$\begin{array}{l}=\cos 45^{\circ} \cos 30^{\circ}-\sin 45^{\circ} \sin 30^{\circ} \\ =\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2} \\ =\frac{\sqrt{3}-1}{2 \sqrt{2}}\end{array}$
Hence, 
$\begin{array}{l}\sin ^2 75^{\circ}-\cos ^2 75^{\circ}=\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)^2-\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^2 \\ =\frac{3+1+2 \sqrt{3}-3-1+2 \sqrt{3}}{8} \\ =\frac{4 \sqrt{3}}{8} \\ =\frac{\sqrt{3}}{2}\end{array}$

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