Question

The value of sin (${45}^{°}+\mathrm{\theta })-\cos ({45}^{°}-\mathrm{\theta })$ is equal to
(a) 2 cos $\mathrm{\theta }$ (b) 0 (c) 2 sin $\mathrm{\theta }$ (d) 1

Answer 1

We know that, $\sin \left(90-\theta \right)=\cos \theta$.
So, $\sin \left(45°+\theta \right)=\cos \left[90-\left(45°+\theta \right)\right]=\cos \left(45°-\theta \right)$
$$\begin{gathered} \therefore \sin \left(45°+\mathrm{\theta }\right)-\cos \left(45°-\mathrm{\theta }\right) \\ =\cos \left(45°-\mathrm{\theta }\right)-\cos \left(45°-\mathrm{\theta }\right) \\ =0 \end{gathered}$$
Hence, the correct answer is option (b).