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    The value of cos (45 + θ) - sin (45° + θ) is:
    Question

    The value of cos (45 + θ) - sin (45° + θ) is:

    A.

    2 sinθ-\sqrt2\space\text{sin}θ​​

    B.

    2sinθ\sqrt{2\text{sin}θ}​​

    C.

    cosθ{\text{cos}θ}​​

    D.

    sinθ\text{sin}θ​​

    Correct option is A

    cos(45+θ)=cos45cosθsin45sinθ=12(cosθsinθ)sin(45+θ)=sin45cosθ+cos45sinθ=12(cosθ+sinθ)cos(45+θ)sin(45+θ)=12(cosθsinθ)12(cosθ+sinθ)=12[(cosθsinθ)(cosθ+sinθ)]=12[cosθsinθcosθsinθ]=12(2sinθ)=2sinθ\begin{aligned}\cos(45^{\circ}+\theta) &= \cos 45^{\circ}\cos\theta - \sin 45^{\circ}\sin\theta \\&= \frac{1}{\sqrt{2}}(\cos\theta - \sin\theta) \\\\\sin(45^{\circ}+\theta) &= \sin 45^{\circ}\cos\theta + \cos 45^{\circ}\sin\theta \\&= \frac{1}{\sqrt{2}}(\cos\theta + \sin\theta) \\\\\cos(45^{\circ}+\theta) - \sin(45^{\circ}+\theta) &= \frac{1}{\sqrt{2}}(\cos\theta - \sin\theta) - \frac{1}{\sqrt{2}}(\cos\theta + \sin\theta) \\&= \frac{1}{\sqrt{2}}[(\cos\theta - \sin\theta) - (\cos\theta + \sin\theta)] \\&= \frac{1}{\sqrt{2}}[\cos\theta - \sin\theta - \cos\theta - \sin\theta] \\&= \frac{1}{\sqrt{2}}(-2\sin\theta) \\&= -\sqrt{2}\sin\theta\end{aligned}​​

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