Question

The value of $\sin \left( {90^\circ - \theta } \right).\cos \theta + \sin \theta .\cos \left( {90^\circ - \theta } \right)$ is:
A) 0
B) 1
C) 2
D) None of these

Answer 1

Hint:
We can use trigonometric identities to expand the difference of angles on both the terms. Then we can substitute the values of the trigonometric ratios at constant angles. Then we can simplify the terms and apply suitable identity to obtain the required solution.

Complete step by step solution:
We need to find the value of the expression $\sin \left( {90 - \theta } \right).\cos \theta + \sin \theta .\cos \left( {90 - \theta } \right)$
Let $I = \sin \left( {90^\circ - \theta } \right).\cos \theta + \sin \theta .\cos \left( {90^\circ - \theta } \right)$
Let ${I_1} = \sin \left( {90^\circ - \theta } \right).\cos \theta \,\,$ and ${I_2} = \sin \theta .\cos \left( {90^\circ - \theta } \right)$ …. (1)
Now consider the 1st term,
 $ \Rightarrow {I_1} = \sin \left( {90^\circ - \theta } \right).\cos \theta \,\,$
We know that $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ . On using this identity, we get
 $ \Rightarrow {I_1} = \left( {\sin 90^\circ \cos \theta - \cos 90^\circ \sin \theta } \right).\cos \theta \,\,$
We know that $\sin 90^\circ = 1$ and $\cos 90^\circ = 0$ on substituting these values, we get,
 $ \Rightarrow {I_1} = \left( {1 \times \cos \theta - 0 \times \sin \theta } \right).\cos \theta \,\,$
On simplification, we get
 $ \Rightarrow {I_1} = \left( {\cos \theta } \right).\cos \theta \,\,$
So, we have
 $ \Rightarrow {I_1} = {\cos ^2}\theta \,\,$ …. (2)
Now we can consider the 2nd term.
 $ \Rightarrow {I_2} = \sin \theta .\cos \left( {90^\circ - \theta } \right)$
We know that $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$ . On using this identity, we get
 $ \Rightarrow {I_2} = \sin \theta .\left( {\cos 90^\circ \cos \theta + \sin 90^\circ \sin \theta } \right)$
We know that $\sin 90^\circ = 1$ and $\cos 90^\circ = 0$ on substituting these values, we get
 $ \Rightarrow {I_2} = \sin \theta .\left( {0 \times \cos \theta + 1 \times \sin \theta } \right)$
On further simplification, we get
 $ \Rightarrow {I_2} = \sin \theta .\left( {0 + \sin \theta } \right)$
So, we get
 $ \Rightarrow {I_2} = {\sin ^2}\theta $ …. (3)
Now we can substitute these in the expression we need to find the value. Then, we get
 $ \Rightarrow I = \sin \left( {90^\circ - \theta } \right).\cos \theta + \sin \theta .\cos \left( {90^\circ - \theta } \right)$
On substituting equation (1), we will obtain
 $ \Rightarrow I = {I_1} + {I_2}$
Now we can substitute equations (2) and (3). So, we get
 $ \Rightarrow I = {\cos ^2}\theta + {\sin ^2}\theta $
We know that ${\cos ^2}\theta + {\sin ^2}\theta = 1$ . On applying this identity, we get
 $ \Rightarrow I = 1$
Therefore, the required value of the given expression is 1.

So, the correct answer is option B.

Note:
Alternate method to solve this problem is given by,
We need to find the value of the expression $\sin \left( {90 - \theta } \right).\cos \theta + \sin \theta .\cos \left( {90 - \theta } \right)$
Let $I = \sin \left( {90 - \theta } \right).\cos \theta + \sin \theta .\cos \left( {90 - \theta } \right)$
We know that $\sin \left( {90 - \theta } \right) = \cos \theta $ and $\cos \left( {90 - \theta } \right) = \sin \theta $ . On substituting these equations, our expression will become
 $I = \cos \theta \times \cos \theta + \sin \theta \times \sin \theta $
We can write them as squares. So, we will get
 $I = {\cos ^2}\theta + {\sin ^2}\theta $
We know that ${\cos ^2}\theta + {\sin ^2}\theta = 1$ . On applying this identity, we get
 $ \Rightarrow I = 1$
Therefore, the required value of the given expression is 1.