Q.Find the value of $\sin(-30^\circ)+\cos(-60^\circ)$.

$\sin(-420^\circ)=-\dfrac{\sqrt{3}}{2}$.

Step 1 — use the odd property $\sin(-\theta)=-\sin\theta$:

$\sin(-420^\circ)=-\sin(420^\circ).$

Step 2 — reduce the angle by subtracting one full turn of $360^\circ$ (co-terminal angle): …

Solving $p\sec50^\circ = \tan50^\circ$ gives $p = \sin50^\circ$, so the answer is option (d).

We are given

$p\sec 50^\circ = \tan 50^\circ \;\Rightarrow\; p = \frac{\tan 50^\circ}{\sec 50^\circ}.$

Write each ratio in terms of sine and cosine:

$\tan 50^\circ = \frac{\sin 50^\circ}{\cos 50^\circ},\qquad \sec 50^\circ = \frac{1}{\cos 50^\circ}.$

Therefore

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