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Question: The value of \[\sin \left( {90 - \theta } \right) \cdot \cos \theta + \sin \theta \cdot \cos \left( ...

The value of sin⁡(90−θ)⋅cos⁡θ+sin⁡θ⋅cos⁡(90−θ)\sin \left( {90 - \theta } \right) \cdot \cos \theta + \sin \theta \cdot \cos \left( {90 - \theta } \right) is equal to
A.0
B.1
C.2
D.None of these

Explanation

Solution

First, we will use the value sin⁡(90−θ)=cos⁡θ\sin \left( {90 - \theta } \right) = \cos \theta and cos⁡(90−θ)=sin⁡θ\cos \left( {90 - \theta } \right) = \sin \theta in the given value and then the property of trigonometric functions,cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1 in the obtained equation to find the required value.

Complete step by step answer:

We are given that sin⁡(90−θ)⋅cos⁡θ+sin⁡θ⋅cos⁡(90−θ)\sin \left( {90 - \theta } \right) \cdot \cos \theta + \sin \theta \cdot \cos \left( {90 - \theta } \right).

Using the value sin⁡(90−θ)=cos⁡θ\sin \left( {90 - \theta } \right) = \cos \theta and cos⁡(90−θ)=sin⁡θ\cos \left( {90 - \theta } \right) = \sin \theta in the given value, we get

⇒cos⁡θ⋅cos⁡θ+sin⁡θ⋅sin⁡θ ⇒cos⁡2θ+sin⁡2θ  \Rightarrow \cos \theta \cdot \cos \theta + \sin \theta \cdot \sin \theta \\\ \Rightarrow {\cos ^2}\theta + {\sin ^2}\theta \\\

Using the property of trigonometric functions,cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1 in the above equation, we get

⇒1 \Rightarrow 1

Hence, option B is correct.

Note: In these types of questions, the key concept to solve this is to learn about the complementary angles of trigonometric ratios. Students need to learn about the basic trigonometric identities to solve such problems.