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Question: Find the value of \(\sin \theta \cos (90 - \theta ) + \cos \theta \sin (90 - \theta ) = \) \( ...

Find the value of sin⁡θcos⁡(90−θ)+cos⁡θsin⁡(90−θ)=\sin \theta \cos (90 - \theta ) + \cos \theta \sin (90 - \theta ) =
A. 0 B. 1 C. 2 D. 32  {\text{A}}{\text{. 0}} \\\ {\text{B}}{\text{. 1}} \\\ {\text{C}}{\text{. 2}} \\\ {\text{D}}{\text{. }}\dfrac{3}{2} \\\

Explanation

Solution

Hint: Here we go through by the relation of complementary angles of trigonometric ratio. As we know the value of cos⁡(90−θ)\cos (90 - \theta )and sin⁡(90−θ)\sin (90 - \theta )by putting their value we get our result.

Complete step-by-step answer:
Here in the question we have to find the value of sin⁡θcos⁡(90−θ)+cos⁡θsin⁡(90−θ)\sin \theta \cos (90 - \theta ) + \cos \theta \sin (90 - \theta )
As we know that by the complementary angles of trigonometric ratio the value of cos⁡(90−θ)=sin⁡θ\cos (90 - \theta ) = \sin \theta andsin⁡(90−θ)=cos⁡θ\sin (90 - \theta ) = \cos \theta .
By putting these values in the question we get,
=sin⁡θ×sin⁡θ+cos⁡θ×cos⁡θ =sin⁡2θ+cos⁡2θ  = \sin \theta \times \sin \theta + \cos \theta \times \cos \theta \\\ = {\sin ^2}\theta + {\cos ^2}\theta \\\
And we also know the trigonometric identity that the value of sin⁡2θ+cos⁡2θ{\sin ^2}\theta + {\cos ^2}\theta is 1.
=1
Hence option B is the correct answer.

Note: - Whenever you get this type of question the key concept to solve this is to learn about the complementary angles of trigonometric ratios and you have to also keep in mind supplementary angles of trigonometric ratios to solve such types of questions. And also learn about the trigonometric identities.