In a triangle ABC, prove that
(a) sin A = sin (B + C)
(b) sin 2A = sin (2B + 2C)
(c) cos A = – cos (A + B)
(d) \[\tan \left( \dfrac{A+B}{2} \right)=\cot \dfrac{C}{2}\]

Hint: We know that in a triangle, \[A+B+C={{180}^{o}}\], we can write it as \[A+B={{180}^{o}}-C\] and take the sin on both sides and use \[\sin \left( {{180}^{o}}-\theta \right)=\sin \theta \]. Similarly, take cos on both sides and use \[\cos \left( {{180}^{o}}-\theta \right)=-\cos \theta \]. Now, divide the initial equation by 2 and take tan on both sides and use \[\tan \left( 90-\theta \right)=\cot \theta \] to prove the desired result.

Complete step-by-step answer:

In this question, we are given a triangle ABC, we have to prove that

(a) sin A = sin (B + C)

(b) sin 2A = sin (2B + 2C)

(c) cos A = – cos (A + B)

(d) \[\tan \left( \dfrac{A+B}{2} \right)=\cot \dfrac{C}{2}\]

First of all, let us draw a triangle ABC.



We know that in a triangle, the sum of the interior angles is \[{{180}^{o}}\]. So, we get,
\[A+B+C={{180}^{o}}.....\left( i \right)\]

(a) By subtracting A from both the sides of the above equation, we get,

\[B+C={{180}^{o}}-A.....\left( ii \right)\]

By taking sin on both the sides of the above equation, we get,

\[\sin \left( B+C \right)=\sin \left( {{180}^{o}}-A \right)\]

We know that \[\sin \left( {{180}^{o}}-\theta \right)=\sin \theta \]. By using this, we get,

\[\sin \left( B+C \right)=\sin A\]

Hence proved.

(b) Now by multiplying 2 on both the sides of the equation (ii), we get,

\[B+C={{180}^{o}}-A\]

\[\Rightarrow 2\left( B+C \right)=2\left( {{180}^{o}}-A \right)\]

\[\Rightarrow 2B+2C={{360}^{o}}-2A\]

Now by taking sin on both the sides of the above equation, we get,

\[\sin \left( 2B+2C \right)=\sin \left( {{360}^{o}}-2A \right)\]

We know that \[\sin \left( {{360}^{o}}-\theta \right)=-\sin \theta \]. By using this, we get,

\[\sin \left( 2B+2C \right)=-\sin \left( 2A \right)\]

\[\sin 2A=-\sin \left( 2B+2C \right)\]

Hence proved.

(c) By subtracting C from both the sides of equation (i), we get,

\[A+B={{180}^{o}}-C\]

By taking cos on both the sides of the above equation, we get,

\[\cos \left( A+B \right)=\cos \left( {{180}^{o}}-C \right)\]

We know that \[\cos \left( {{180}^{o}}-\theta \right)=-\cos \theta \]. By using this in the above equation, we get,

\[\cos \left( A+B \right)=-\cos C\]

\[\cos C=-\cos \left( A+B \right)\]

Hence proved.

(d) Now by subtracting C from both the sides of the equation (i), we get,

\[A+B={{180}^{o}}-C\]

By dividing by 2 on both the sides of the above equation, we get,

\[\dfrac{A+B}{2}=\dfrac{{{180}^{o}}-C}{2}\]

\[\Rightarrow \dfrac{A+B}{2}=\dfrac{{{180}^{o}}}{2}-\dfrac{C}{2}\]

\[\Rightarrow \dfrac{A+B}{2}={{90}^{o}}-\dfrac{C}{2}\]

By taking tan on both the sides of the above equation, we get,

\[\tan \left( \dfrac{A+B}{2} \right)=\tan \left( {{90}^{o}}-\dfrac{C}{2} \right)\]

We know that \[\tan \left( {{90}^{o}}-\theta \right)=\cot \theta \]. By using this in the above equation, we get,

\[\tan \left( \dfrac{A+B}{2} \right)=\cot \dfrac{C}{2}\]

Hence proved.

Note: Students must note that sum of the interior angles of the triangle is \[{{180}^{o}}\] and the sum of the interior angles of any n sided polygon is \[\left( n-2 \right){{.180}^{o}}\]. In these types of questions, students should properly read the question and then transpose one angle to another to prove the desired result. Also, students must keep in mind that \[\sin \left( {{180}^{o}}-\theta \right)=\sin \theta \] while \[\cos \left( {{180}^{o}}-\theta \right)=-\cos \theta \] and should not get confused between the two.