If in A B C, 2 b^2=a^2+c^2, then 3 B B = - Vidyadip

MCQ

If in $\triangle A B C, 2 b^2=a^2+c^2$, then $\frac{\sin 3 B}{\sin B}=$

A
$\frac{ c ^2- a ^2}{2 ca }$
B
$\frac{c^2-a^2}{c a}$
C
$\left(\frac{ c ^2- a ^2}{ ca }\right)^2$
✓
$\left(\frac{ c ^2- a ^2}{2 ca }\right)^2$

✓

Answer

Correct option: D.

$\left(\frac{ c ^2- a ^2}{2 ca }\right)^2$

(D) $\frac{\sin 3 B}{\sin B}=\frac{3 \sin B-4 \sin ^3 B}{\sin B}=3-4 \sin ^2 B$
$=3-4+4 \cos ^2 B$
$=-1+\frac{4\left(a^2+c^2-b^2\right)^2}{4(a c)^2}$
$=-1+\frac{\left(\frac{a^2+c^2}{2}\right)^2}{(a c)^2} \ldots\left[\because 2 b^2=a^2+c^2\right]$
$=-1+\frac{\left( a ^2+ c ^2\right)^2}{4( ac )^2}$
$=\frac{\left( a ^2+ c ^2\right)^2-4 a ^2 c ^2}{4( ac )^2}$
$=\left(\frac{ c ^2- a ^2}{2 ac }\right)^2$

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 "MCQ" from Trigonometric Functions→Trigonometric Functions — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

The line joining the points (-2, 1, -8) and (a, b, c) is parallel to the line whose direction ratios are 6, 2, 3. The value of a, b, c are

If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two collinear vectors, then which of the follwoing are incorrect?

The equation of line passing through $(3,-1,2)$ and perpendiculartothelines $\vec{r}=(\hat{i}+\hat{j}-\hat{k})+ l (2 \hat{i}-2 \hat{j}+\hat{k})$ and $\vec{r}=(2 \hat{i}+\hat{j}-3 \hat{k})+ m (\hat{i}-2 \hat{j}+2 \hat{k})$ is

If $D , E , F$ be the middle points of the sides BC , CA and AB of the triangle ABC , then $\overline{ AD }+\overline{ BE }+\overline{ CF }$ is

Which of the following statement is contingency?

Forces $3\overrightarrow{\text{OA}},\ 5\overrightarrow{\text{OB}}$ act along $\text{OA}$ and $\text{OB}$. If their resultant passes through $\text{C}$ on $\text{AB}$, then,

If $\tan ^{-1} x+2 \cot ^{-1} x=\frac{2 \pi}{3}$, then $x=$

If $\cos ^{-1}\left(\frac{1}{x}\right)=\theta$, then $\tan \theta=$

The Solution of $\frac{d y}{d x}=\frac{y+\sqrt{x^2-y^2}}{2}$ is......

If $\cos ^{-1} \frac{3}{5}-\sin ^{-1} \frac{4}{5}=\cos ^{-1} x$, then $x=$