Question

Show that $\cos \theta +sin({270}^o+\theta )-\sin ({270}^o-\theta )+\cos ({180}^o+\theta )=0$

Answer 1

L.H.S

$=\cos \theta +\sin ({270}^{\circ }+\theta )-\sin ({270}^{\circ }+\theta )+\cos ({180}^{\circ }+\theta )$

We know that

$\sin ({270}^{\circ }+\theta )=-\cos \theta$

$\cos ({180}^{\circ }+\theta )=-\cos \theta$

Therefore,

$=\cos \theta +(-\cos \theta )-(-\cos \theta )+(-\cos \theta )$

$=\cos \theta -\cos \theta +\cos \theta -\cos \theta$

$=0$

Hence, this is the answer.