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

Hint: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 \[\cos \left( {{90}^{\circ }}-A \right)=\sin A,\sin \left( {{90}^{\circ }}-A \right)=\cos A\].