Question

Find the value of $\cos \left( {{40}^{\circ }}+\theta \right)-\sin \left( {{50}^{\circ }}-\theta \right)$ ?
A. 1
B. 0
C. $\sin {{20}^{\circ }}$
D. None of these

Answer 1

First let us assume that the value of $\left( {{40}^{\circ }}+\theta \right)$ as $x$ after that we change the ${{50}^{\circ }}$ into ${{40}^{\circ }}$ by subtracting it with ${{90}^{\circ }}$ thereby finding the value of the given term.

Complete step by step solution:
According to the question given, we assume that the angle $\left( {{40}^{\circ }}+\theta \right)$ is equal to $x$ and after that we place the value in the equation $\cos \left( {{40}^{\circ }}+\theta \right)- \sin \left( {{50}^{\circ }}-\theta \right)$.
So let us place the value in the above term as:
$\Rightarrow \cos \left( x \right)-\sin \left( {{50}^{\circ }}-\theta \right)$
After this we change the value of $\sin \left( {{50}^{\circ }}-\theta \right)$ into such a form that we can place the value $x$ in it so we take ${{90}^{\circ }}$ and subtract it by ${{50}^{\circ }}$ to get:
$\Rightarrow \cos \left( x \right)-\sin \left( {{90}^{\circ }}-x \right)$ (Placing the value of $\left( {{40}^{\circ }}+\theta \right)$ is equal to $x$)
Now according to trigonometry identities, the value of $\sin \left( {{90}^{\circ }}-A \right)=\cos A$, we get the value as:
$\Rightarrow \cos \left( x \right)-\cos \left( x \right)=0$
Therefore, the value of $\cos \left( {{40}^{\circ }}+\theta \right)-\sin \left( {{50}^{\circ }}-\theta \right)=0$.

Note: The value of the trigonometry identity is given as $\cos \left( {{90}^{\circ }}-A \right)=\sin A,\sin \left( {{90}^{\circ }}-A \right)=\cos A$.