Question

If A, B, C, D be the angles of a cyclic quadrilateral, taken in order, prove that
\[cos\left( 180-A \right)+cos\left( 180+B \right)+cos\left( 180+C \right)-sin\left( 90+D \right)=0\]

Answer 1

Hint: -In such question, we prove them by either making the left hand side that is L.H.S. or by making the right hand side that is R.H.S. equal to the other in order to prove the proof that has been asked.

Complete step-by-step answer:
Some important formulae that might be used in solving this question is as follows
\[\begin{align}
  & \sin (-x)=-\sin x \\
 & \sin ({{90}^{\circ }}-x)=\cos x \\
 & \sin ({{90}^{\circ }}+x)=\cos x \\
 & \sin ({{180}^{\circ }}-x)=\sin x \\
 & \sin ({{180}^{\circ }}+x)=-\sin x\ \ \ \ \ \\
 & \\
 & \cos (-x)=\cos x \\
 & \cos ({{90}^{\circ }}-x)=\sin x \\
 & \cos ({{90}^{\circ }}+x)=-\sin x \\
 & \cos ({{180}^{\circ }}-x)=-\cos x \\
 & \cos ({{180}^{\circ }}+x)=-\cos x \\
\end{align}\]
Now, these are the results that would be used to prove the proof mentioned in this question as using these identities, we would convert the left hand side that is L.H.S. or the right hand side that is R.H.S. to make either of them equal to the other.
Also, another important fact or information that would be used in solving this question is that the sum of opposite angles of a cyclic quadrilateral is equal to \[{{180}^{\circ }}\] .
In this particular question, we will change the trigonometric functions according to the formulae that are mentioned above, then we will use the information regarding the cyclic quadrilateral and so we will try to get to a solution.

As mentioned in the question, we have to prove the given expression.
Now, we will start with the left hand side that is L.H.S. and try to make the necessary changes that are given in the hint, first, and then on simplifying the angles of the trigonometric functions, we get the following result
\[=-\cos (A)-\cos (B)-\cos (C)-\cos (D)\ \ \ \ \ ...(b)\]
(Using the identities that are mentioned in the hint)
Now, as mentioned already, the sum of the opposite angles of a cyclic quadrilateral is \[{{180}^{\circ }}\] , hence, we can write the follows
\[\begin{align}
  & A+C={{180}^{\circ }}\ and\ B+D={{180}^{\circ }} \\
 & A={{180}^{\circ }}-C\ and\ B={{180}^{\circ }}-D\ \ \ \ \ ...(a) \\
\end{align}\]
Now, on using equation (a) in the equation (b), we get the following result
\[\begin{align}
  & =-\cos ({{180}^{\circ }}-C)-\cos ({{180}^{\circ }}-D)-\cos (C)-\cos (D) \\
 & =-\left( -\cos (C) \right)-\left( -\cos (D) \right)-\cos (C)-\cos (D) \\
 & =\cos (C)+\cos (D)-\cos (C)-\cos (D) \\
\end{align}\]
Now, on cancellation of terms, we get the following
\[=0\]

Now, as the right hand side that is R.H.S. is equal to the left hand side that is L.H.S., hence, the expression has been proved.

Note: Another method of attempting this question is by converting the right hand side that is R.H.S. to the left hand side that is L.H.S. by using the relations that are given in the hint. Through this method also, we could get to the correct answer and hence, we would be able to prove the required proof.