Download App

Model Paper 5 — Maths STD 11 Science — Question

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

Answer

Correct option: A.
$\sqrt{3} / 2$
$\sin ^2 75^{\circ}-\sin ^2 15^{\circ}$
$=\sin ^2 75^{\circ}-\cos ^2 75^{\circ}\left[\sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$
Now, $\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)$
$=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}$
$=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}$
$=\frac{\sqrt{3}+1}{2 \sqrt{2}}$
$\cos 75^{\circ}=\cos \left(45^{\circ}+30^{\circ}\right)$
$=\cos 45^{\circ} \cos 30^{\circ}-\sin 45^{\circ} \sin 30^{\circ}$
$=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}$
$=\frac{\sqrt{3}-1}{2 \sqrt{2}}$
Hence, $\sin ^2 75^{\circ}-\cos ^2 75^{\circ}$
$=\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)^2-\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^2$
$=\frac{3+1+2 \sqrt{3}-3-1+2 \sqrt{3}}{8}$
$=\frac{4 \sqrt{3}}{8}$
$=\frac{\sqrt{3}}{2}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Model Paper 5→Model Paper 5 — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $ is equal to
→
Let $S$ denotes the set of all real values of the parameter $'a\ '$ for which every solution of the inequality $\log\frac{1}{2} \text{x}^2\geq \log\frac{1}{2} (\text{x} + 2)$ is the solution of the inequality $49\text{x}^2-4\text{a}^4\leq0$ What is the value of $S?$
→
The solution of inequality $x^2-x-6<0$ is :
→
$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^3} + 1}} + \frac{4}{{{n^3} + 1}} + \frac{9}{{{n^3} + 1}} + ........ + \frac{{{n^2}}}{{{n^3} + 1}}} \right] = $
→
The arithmetic mean and the geometric mean of two distinct 2-digit numbers $x$ and $y$ are two integers one of which can be obtained by reversing the digits of the other (in base 10 representation). Then, $x+y$ equals
→
Let $P(m)$ be the statement $m^2 > 100$, the statement $P(k + 1)$ will be true if:
→
If $10^{3n}+ 2^{4k+1}× 9 + k,$ is divisible by $11,$ then what is the least positive value of $k?$
→
What is the value of the $\lim_\limits{\text{x} \rightarrow 5}\frac{32\text{x}+1}{\text{x}^2-5\text{x}}?$
→
The contrapositive of $2x + 3 = 9 \Rightarrow x \neq 4$ is.
→
If $tan\ 80^o = a$ and $tan47^o = b$, then $tan37^o$ is equal to -
→