Maths Questions

Solution : The general solution of $cos \theta$ = $cos \alpha$ is given by $\theta$ = $2n\pi \pm \alpha$,  n $\in$ Z. Proof : We have,  $cos \theta$ = $cos \alpha$ $\implies$  $cos \theta$ – $cos \alpha$ = 0 $\implies$   -$2 sin ({\theta + \alpha\over 2}) sin({\theta – \alpha\over 2})$ = 0 $\implies$  \(sin ({\theta […]

Solution : The general solution of $sin \theta$ = $sin \alpha$ is given by $\theta$ = $n\pi + (-1)^n \alpha$,  n $\in$ Z. Proof : We have,  $sin \theta$ = $sin \alpha$ $\implies$  $sin \theta$ – $sin \alpha$ = 0 $\implies$   $2 sin ({\theta – \alpha\over 2}) cos({\theta + \alpha\over 2})$ = 0 $\implies$  \(sin

Solution : We have, $\sqrt{3} cos \theta$ + $sin \theta$ = $\sqrt{2}$            ………….(i) This is of the form $a cos\theta$ + $b sin \theta$ = c, where a = $\sqrt{3}$, b = 1 and c = $\sqrt{2}$. Let a = $r cos\alpha$ and b = $r sin\alpha$. Then, $\sqrt{3}$ =

Solution : The general solution of $sin^2 \theta$ = $sin^2 \alpha$ is given by $\theta$ = $n\pi \pm \alpha$, n $\in$ Z. Proof : We have, $sin^2 \theta$ =$sin^2 \alpha$ $\implies$  $2 sin^2 \theta$ =$2 sin^2 \alpha$ $\implies$  $1 – cos 2\theta$ = $1 – cos 2\alpha$ $\implies$   $cos 2\theta$ = $cos 2\alpha$ $\implies$  $2\theta$

Solution : The general solution of $cos^2 \theta$ = $cos^2 \alpha$ is given by $\theta$ = $n\pi \pm \alpha$, n $\in$ Z. Proof : We have, $cos^2 \theta$ =$cos^2 \alpha$ $\implies$  $2 cos^2 \theta$ =$2 cos^2 \alpha$ $\implies$  $1 + cos 2\theta$ = $1 + cos 2\alpha$ $\implies$   $cos 2\theta$ = $cos 2\alpha$ $\implies$  $2\theta$

Solution : The general solution of $tan^2 \theta$ = $tan^2 \alpha$ is given by $\theta$ = $n\pi \pm \alpha$, n $\in$ Z. Proof : We have, $tan^2 \theta$ =$tan^2 \alpha$ $\implies$  $1 – tan^2\theta\over 1 + tan^2 \theta$ =$1 – tan^2\alpha\over 1 + tan^2 \alpha$ $\implies$   $cos 2\theta$ = $cos 2\alpha$ $\implies$  $2\theta$ = \(2n\pi

Solution : The general solution of $cot \theta$ = 0 is given by $\theta$ = $(2n + 1){\pi\over 2}$, n $\in$ Z. Proof : We have, $cot \theta$ = $OM\over PM$ $\therefore$   $cot \theta$ = 0 $\implies$  $OM\over PM$ = 0 $\implies$ OM = 0 $\implies$  OP coincides with OY or OY’ $\implies$  $\theta$ =

Solution : The general solution of $cos \theta$ = 0 is given by $\theta$ = $(2n + 1){\pi\over 2}$, n $\in$ Z. Proof : We have, $cos \theta$ = $PM\over OP$ $\therefore$   $cos \theta$ = 0 $\implies$  $OM\over OP$ = 0 $\implies$ OM = 0 $\implies$  OP coincides with OY or OY’ $\implies$  $\theta$ = 

Solution : The general solution of $tan \theta$ = 0 is given by $\theta$ = $n\pi$, n $\in$ Z. Proof : We have, $tan \theta$ = $PM\over OM$ $\therefore$   $tan \theta$ = 0 $\implies$  $PM\over OM$ = 0 $\implies$ PM = 0 $\implies$  OP coincides with OX or OX’ $\implies$  $\theta$ = 0, $\pi$, $2\pi$,

Solution : The general solution of $sin \theta$ = 0 is given by $\theta$ = $n\pi$, n $\in$ Z. Proof : We have, $sin \theta$ = $PM\over OP$ $\therefore$   $sin \theta$ = 0 $\implies$  $PM\over OP$ = 0 $\implies$ PM = 0 $\implies$  OP coincides with OX or OX’ $\implies$  $\theta$ = 0, $\pi$, $2\pi$,