જો $x$ અને $y$ બંને બીજા ચરણમાં હોય અને $\sin x=\frac{3}{5}, \cos y=-\frac{12}{13},$ તો $\sin (x+y)$ નું મૂલ્ય શોધો.
We know that
$\sin (x+y)=\sin x \cos y+\cos x \sin y$.......$(1)$
Now $\cos ^{2} x=1-\sin ^{2} x=1-\frac{9}{25}=\frac{16}{25}$
Therefore $\cos x=\pm \frac{4}{5}$
since $x$ lies in second quadrant, cos $x$ is negative.
Hence $\cos x=-\frac{4}{5}$
Now $\sin ^{2} y=1-\cos ^{2} y=1-\frac{144}{169}=\frac{25}{169}$
i.e. $\sin y=\pm \frac{5}{13}$
since $y$ lies in second quadrant, hence sin $y$ is positive. Therefore, $\sin y=\frac{5}{13} .$ Substituting the values of $\sin x, \sin y, \cos x$ and $\cos y$ in $(1),$ we get
$\sin (x+y)=\frac{3}{5} \times\left(-\frac{12}{13}\right)+\left(-\frac{4}{5}\right) \times \frac{5}{13}$
$\frac{36}{65}-\frac{20}{65}=-\frac{56}{65}$
જો ${\sec ^2}\theta = \frac{4}{3}$, તો $\theta $ નો વ્યાપક ઉકેલ મેળવો.
જો $\left| {\,\begin{array}{*{20}{c}}{\cos (A + B)}&{ - \sin (A + B)}&{\cos 2B}\\{\sin A}&{\cos A}&{\sin B}\\{ - \cos A}&{\sin A}&{\cos B}\end{array}\,} \right| = 0$ તો $B =$
જો $\tan \theta + \tan 2\theta + \sqrt 3 \tan \theta \tan 2\theta = \sqrt 3 ,$ તો
જો $sin^2x + sinx \,cosx -6cos^2x = 0$ અને $-\frac{\pi}{2} < x < 0$,હોય તો $cos2x$ ની કિમત મેળવો.
$2{\sin ^2}x + {\sin ^2}2x = 2,\, - \pi < x < \pi ,$ તો $x = $