Download App

Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
MCQ
$\sin 15^{\circ}+\cos 105^{\circ}=$
  • ✓
    $0$
  • B
    $2 \sin 15^{\circ}$
  • C
    $\cos 15^{\circ}+\sin 15^{\circ}$
  • D
    $\sin 15^{\circ}-\cos 15^{\circ}$
✓

Answer

Correct option: A.
$0$
(A)
$\sin 15^{\circ}+\cos 105^{\circ}$
$=\sin 15^{\circ}+\cos \left(90^{\circ}+15^{\circ}\right)$
$=\sin 15^{\circ}-\sin 15^{\circ}$
$\ldots\left[\because \cos \left(90^{\circ}+\theta\right)=-\sin \theta\right]$
= 0

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 "MCQ" from Trigonometry – II→Trigonometry – II — all question types→Maths STD 11 Science question papers→Maharashtra Board STD 11 Science question papers→Maharashtra Board question paper generator→

Similar questions

$\lim _{x \rightarrow 1} \frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}=$
→
Two tangents PQ and PR drawn to the circle $x^2+y^2-2 x-4 y-20=0$ from point P(16, 7). If the centre of the circle is C, then the area of quadrilateral PQCR will be
→
If pth, qth and rth terms of an $A.P.$ are in $G.P.$, then the common ratio of this $G.P.$ is:
→
The equation of the circle whose radius is 5 and which touches the circle $x^2+y^2-2 x-4 y-20=0$ externally at the point (5, 5), is
→
The intercept cut off from Y-axis is twice that from X-axis by the line and line passes through (1, 2), then its equation is
→
If $\frac{-\pi}{2}<\theta<\frac{\pi}{2}$ and $\theta \neq \pm \frac{\pi}{4}$, then the value of $\cot \left(\frac{\pi}{4}+\theta\right) \cot \left(\frac{\pi}{4}-\theta\right)$ is
→
The locus of the mid-points of the perpendiculars drawn from points on the line, x = 2y to the line x = y is
→
The value of $2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-16\cos^3\text{x}\sin^2\text{x}$ is :
→
$\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=$
→
If $\alpha$ and $\beta$ are imaginary cube roots of unity, then the value of $\alpha^4+\beta^{28}+\frac{1}{\alpha \beta}$ is
→