Find the angle in radian through which a pendulum swings if its length is $75\, cm$ and the tip describes an arc of length.
$10 \,cm$
We know that in a circle of radius $r$ unit, if an arc of length $l$ unit subtends
An angle $\theta$ radian at the centre, then $\theta=\frac{l}{r}$
It is given that $r=75 \,cm$
Here, $l=10\, cm$
$\theta=\frac{10}{75}\, radian =\frac{2}{15}\, radian$
If $p = \frac{{2\sin \,\theta }}{{1 + \cos \theta + \sin \theta }}$, and $q = \frac{{\cos \theta }}{{1 + \sin \theta }},$ then
If $\left| {\,a\,{{\sin }^2}\theta + b\sin \theta \cos \theta + c\,{{\cos }^2}\theta - \frac{1}{2}(a + c)\,} \right|\, \le \frac{1}{2}k,$ then ${k^2}$ is equal to
The value of $\cos A - \sin A$ when $A = \frac{{5\pi }}{4},$ is
Let the function $:(0, \pi) \rightarrow R$ be defined by
$f (\theta)=(\sin \theta+\cos \theta)^2+(\sin \theta-\cos \theta)^4$
Suppose the function $f$ has a local minimum at $\theta$ precisely when $\theta \in\left\{\lambda_1 \pi, \ldots, \lambda_{ T } \pi\right\}$, where $0<\lambda_1<\cdots<\lambda_r<1$. Then the value of $\lambda_1+\cdots+\lambda_r$ is. . . . .
If $(\sec \alpha + \tan \alpha )(\sec \beta + \tan \beta )(\sec \gamma + \tan \gamma )$$ = \tan \alpha \tan \beta \tan \gamma $, then $(\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )$$(\sec \gamma - \tan \gamma ) = $