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ID:
Class: 11
Subject: Math
Topic: Trigonometric Functions
Type: Short
Question:
Prove that: \( \cot^2\frac{\pi}{6} + \text{cosec}\frac{5\pi}{6} + 3\tan^2\frac{\pi}{6} = 6. \)
Official Solution
Explanation:
We know the standard values:
\[
\cot\frac{\pi}{6} = \sqrt{3}, \quad \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}.
\]
Also,
\[
\text{cosec}\left(\frac{5\pi}{6}\right) = \text{cosec}\left(\pi - \frac{\pi}{6}\right) = \text{cosec}\left(\frac{\pi}{6}\right) = 2.
\]
Now substituting these values in the LHS,
\[
\text{LHS} = \cot^2\frac{\pi}{6} + \text{cosec}\frac{5\pi}{6} + 3\tan^2\frac{\pi}{6}
\]
\[
= (\sqrt{3})^2 + 2 + 3\left(\frac{1}{\sqrt{3}}\right)^2
\]
\[
= 3 + 2 + 3\cdot \frac{1}{3}
\]
\[
= 3 + 2 + 1 = 6.
\]
\[
\therefore \text{LHS} = \text{RHS}.
\]
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