Question

Prove that $\cos A + \sin (270 - A) - \sin (270 - A) + \cos (180 + A) = 0$

Answer 1

Hint: As we know that if the angle say $\theta $ is in the third quadrant then $\cos \theta $ will be negative and we know that $\sin (A + B) = \sin A\cos B + \sin B\cos A$. We are going to use this formula to get the required answer.

Complete step by step answer:
Here we are given the equation in terms of sin and cos and we need to prove that
$\cos A + \sin (270 - A) - \sin (270 - A) + \cos (180 + A) = 0$
Now if we take LHS
So LHS$ = \cos A + \sin (270 - A) - \sin (270 - A) + \cos (180 + A)$$ - - - - (1)$
So we need to take LHS
As we know that for the given angle $A,B$
$\sin (A + B) = \sin A\cos B + \sin B\cos A$
So we can find the value of $\sin (270 + A)$
$\Rightarrow \sin (270 + A)$$ = \sin 270\cos A + \cos 270\sin A$
We know that
$\sin (2n + 1)\dfrac{\pi }{2} = {( - 1)^n}$
So for $n = 1$
$\sin \dfrac{{3\pi }}{2} = {( - 1)^1}$
And we know that $\dfrac{{3\pi }}{2} = 270^\circ $
So $\sin (270^\circ ) = - 1$
And also we know that $\cos (2n + 1)\dfrac{\pi }{2} = 0$
Hence $\cos (270^\circ ) = 0$
Therefore we can say that
$\Rightarrow \sin (270 + A)$$ = \sin 270\cos A + \cos 270\sin A$
$
   = ( - 1)\cos A + 0.\sin A \\
   = - \cos A \\
 $
And we also know that $\sin (A - B) = \sin A\cos B - \sin B\cos A$
Now we get that
$\Rightarrow \sin (270 - A)$$ = \sin 270\cos A - \cos 270\sin A$
$
   = - \cos A - 0.\sin A \\
   = - \cos A
$
Therefore we get that
$\Rightarrow \sin (270 + A)$$ = - \cos A$
$\Rightarrow \sin (270 - A) = - \cos A$
Also we know that we need to find$\cos (180 + A)$ and we know that
As $180 + A$ is in the third quadrant, then we know that here cos is negative
So $\cos (180 + A) = - \cos A$
Now putting the value in equation (1) we get that
$\Rightarrow LHS = \cos A + \sin (270 - A) - \sin (270 - A) + \cos (180 + A)$
$
   = \cos A - \cos A + \cos A - \cos A \\
   = 0 \\
= RHS
 $

Hence proved.

Note:
We know that $\sin (90 + \theta ) = \cos \theta $
Similarly for $\sin \left( {\dfrac{{3\pi }}{2} + \theta } \right) = - \cos \theta $ as $\dfrac{{3\pi }}{2}$ lies in the fourth quadrant. For the first quadrant all are positive, for the second, sin and cosec are positive, for the third, tan and cot are positive while in the fourth quadrant cos and sec are positive.