MCQ
If $\tan \theta = \frac{{ - 4}}{3},$ then $\sin \theta = $
  • A
    $-4/5$ but not $4/5$
  • $-4/5 $ or $4/5$
  • C
    $4/5$ but not $-4/5$
  • D
    None of these

Answer

Correct option: B.
$-4/5 $ or $4/5$
b
(b) Since ${\rm{cose}}{{\rm{c}}^2}\theta = 1 + {\cot ^2}\theta = 1 + \frac{9}{{16}} = \frac{{25}}{{16}}$

$\left( \because   {\tan \theta  =  - \frac{4}{3}} \right)$

${\sin ^2}\theta = \frac{1}{{{\rm{cose}}{{\rm{c}}^2}\theta }} = \frac{{16}}{{25}} $

$\Rightarrow \sin \theta = \pm \frac{4}{5},$ 

Both the values are acceptable, since $\tan \theta = - \frac{4}{3}\,\,$ 

$\,i.e.,\theta $ lies in ${2^{nd}}$ or ${4^{th}}$ quadrant.

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

Similar questions

If equation of line is $y = 5x + 10$ then find the value of $x-$intercept made by the line:
$\sum_{r=0}^n 4^r \cdot{ }^n C_r$ is equal to
If 1, $\omega ,\,{{\omega }^{2}}$ are the cube roots of unity then ${{\omega }^{2}}{{(1+\omega )}^{3}}-(1+{{\omega }^{2}})\omega =$ [Orissa JEE 2005]
The least value of $\alpha \in \mathbb{R}$ for which $4 \alpha x^2+\frac{1}{x} \geq 1$, for all $x>0$, is
For the four circles $M , N , O$ and $P ,$ following four equations are given

Circle $M : x ^{2}+ y ^{2}=1$ ; Circle $N : x ^{2}+ y ^{2}-2 x =0$ ; Circle $O : x ^{2}+ y ^{2}-2 x -2 y +1=0$ ;Circle $P: x^{2}+y^{2}-2 y=0$

If the centre of circle $M$ is joined with centre of the circle $N$, further centre of circle $N$ is joined with centre of the circle $O ,$ centre of circle $O$ is joined with the centre of circle $P$ and lastly, centre of circle $P$ is joined with centre of circle $M ,$ then these lines form the sides of a

If $\alpha, \beta$ are the distinct roots of $x^{2}+b x+c=0$ then $\lim _{x \rightarrow \beta} \frac{e^{2\left(x^{2}+b x+c\right)}-1-2\left(x^{2}+b x+c\right)}{(x-\beta)^{2}}$ is equal to:
$^{80}C_{40 }$ is not divisible by -
If the coefficients of ${x^7}$ and ${x^8}$ in ${\left( {2 + \frac{x}{3}} \right)^n}$ are equal, then $n$ is
If $|x|<1,|y|<1$ and $x \neq y,$ then the sum to infinity of the following series $(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots .$
Kavita obtained $16, 14, 18$ and $20$ marks $($out of $25)$ in maths in weekly test in the month of Jan $2000;$ then mean marks of Kavita is: