Q.Prove that $3\sin\frac{\pi}{6}\sec\frac{\pi}{3} - 4\sin\frac{5\pi}{6}\cot\frac{\pi}{4} = 1$.

Concept: Trigonometric values in different quadrants — specifically, knowing that $\sin\frac{5\pi}{6} = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\frac{\pi}{6}$.

Step 1: Evaluate each standard angle.

$\sin\frac{\pi}{6} = \frac{1}{2}$, $\sec\frac{\pi}{3} = \frac{1}{\cos\frac{\pi}{3}} = \frac{1}{1/2} = 2$, $\cot\frac{\pi}{4} = 1$.

Step 2: For $\sin\frac{5\pi}{6}$, note $\frac{5\pi}{6}$ lies in Quadrant II where sine is positive, and its reference angle is $\frac{\pi}{6}$. Hence $\sin\frac{5\pi}{6} = \sin\frac{\pi}{6} = \frac{1}{2}$.

Step 3: Substitute into the expression:

$3 \cdot \frac{1}{2} \cdot 2 - 4 \cdot \frac{1}{2} \cdot 1 = 3 \cdot 1 - 2 = 3 - 2 = 1$.

This problem tests your ability to handle trigonometric functions at standard angles — angles like $\frac{\pi}{6}$, $\frac{\pi}{3}$, $\frac{5\pi}{6}$, and $\frac{\pi}{4}$ appear frequently in exams. The trick is not to rush: evaluate each term separately, paying close attention to the sign of the function in the correct quadrant.

Let’s break it down step by step.

1. Evaluate $\sin\frac{\pi}{6}$

   $\frac{\pi}{6}$ is $30^\circ$, in the first quadrant where sine is positive.

   $\sin\frac{\pi}{6} = \frac{1}{2}$.

2. Evaluate $\sec\frac{\pi}{3}$

   $\frac{\pi}{3}$ is $60^\circ$, also in the first quadrant.

   $\sec\theta = \frac{1}{\cos\theta}$, and $\cos\frac{\pi}{3} = \frac{1}{2}$.

   So $\sec\frac{\pi}{3} = \frac{1}{1/2} = 2$.

   Tip

   Remember: $\sec$ and $\cos$ are reciprocals. If you know $\cos 60^\circ = \frac12$, then $\sec 60^\circ = 2$ instantly.

3. Evaluate $\sin\frac{5\pi}{6}$

   $\frac{5\pi}{6}$ is $150^\circ$. This angle lies in the second quadrant, where sine is positive (only sine and cosecant are positive there).

   The reference angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$ (i.e., $30^\circ$).

   So $\sin\frac{5\pi}{6} = \sin\frac{\pi}{6} = \frac{1}{2}$.

   Watch out

   A common mistake is to think $\sin\frac{5\pi}{6}$ is negative because the angle looks large. But in the second quadrant, sine is positive — only cosine and tangent are negative there.

4. Evaluate $\cot\frac{\pi}{4}$

   $\frac{\pi}{4}$ is $45^\circ$, in the first quadrant.

   $\cot\theta = \frac{1}{\tan\theta}$, and $\tan\frac{\pi}{4} = 1$.

   So $\cot\frac{\pi}{4} = 1$.

5. Now substitute into the expression

   The given expression is:

Plug in the values:

6. Simplify term by term First term: $3 \cdot \frac12 \cdot 2 = 3 \cdot 1 = 3$ Second term: $4 \cdot \frac12 \cdot 1 = 2$ So the expression becomes: