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Find the value of $\cos (270^{\circ} + θ) \cdot \cos (90^{\circ} - θ) - \sin (270^{\circ} - θ) \cdot \cos θ$

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- 1

- \frac{1}{2}

answer is D.

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Detailed Solution

Given,
The trigonometric expression,

\cos (270^{\circ} + θ) \cos (90 - θ) - \sin (270 - θ) \cos θ

It is known that,

\cos (270^{\circ} + θ) = \sin θ

\cos (90 - θ) = \sin θ

and

\sin (270 - θ) = - \cos θ

Put these values in given expression,

\cos (270^{\circ} + θ) \cos (90 - θ) - \sin (270 - θ) \cos θ

= \sin θ \cdot \sin θ - (- \cos θ) \cos θ

= \sin^{2} θ + \cos^{2} θ

= 1

(We now

\sin^{2} θ + \cos^{2} θ = 1

)

Thus,

\cos (270^{\circ} + θ) \cos (90 - θ) - \sin (270 - θ) \cos θ = 1

Thus, the correct option is 4.

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