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Exercise 3.3 · Q3

Q.Prove that cot⁡2π6+csc⁡5π6+3tan⁡2π6=6\cot^2\frac{\pi}{6} + \csc\frac{5\pi}{6} + 3\tan^2\frac{\pi}{6} = 6.

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The key idea is to evaluate each trigonometric term using standard angle values and quadrant signs, then simplify the expression to show it equals 6.

We need to prove that cot⁡2π6+csc⁡5π6+3tan⁡2π6=6\cot^2\frac{\pi}{6} + \csc\frac{5\pi}{6} + 3\tan^2\frac{\pi}{6} = 6. This is a direct verification problem — we compute each term separately using known values of trigonometric functions at standard angles, paying careful attention to the quadrant for 5π6\frac{5\pi}{6}.

The angle π6\frac{\pi}{6} (30°) is in the first quadrant, where all trigonometric ratios are positive. The angle 5π6\frac{5\pi}{6} (150°) is in the second quadrant, where sine is positive but cosine is negative — this affects cosecant, which is the reciprocal of sine.

Let’s work through it step by step.

  1. Evaluate cot⁡π6\cot\frac{\pi}{6} and tan⁡π6\tan\frac{\pi}{6}

    We know tan⁡π6=13\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}. Therefore, cot⁡π6=1tan⁡π6=3\cot\frac{\pi}{6} = \frac{1}{\tan\frac{\pi}{6}} = \sqrt{3}.

    So cot⁡2π6=(3)2=3\cot^2\frac{\pi}{6} = (\sqrt{3})^2 = 3.

  2. Evaluate tan⁡2π6\tan^2\frac{\pi}{6}

    tan⁡π6=13\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}, so tan⁡2π6=(13)2=13\tan^2\frac{\pi}{6} = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}.

  3. Evaluate csc⁡5π6\csc\frac{5\pi}{6}

    5π6\frac{5\pi}{6} is in the second quadrant. The reference angle is π−5π6=π6\pi - \frac{5\pi}{6} = \frac{\pi}{6}.

    In the second quadrant, sin⁡θ\sin\theta is positive, so sin⁡5π6=sin⁡π6=12\sin\frac{5\pi}{6} = \sin\frac{\pi}{6} = \frac{1}{2}.

    Hence csc⁡5π6=1sin⁡5π6=11/2=2\csc\frac{5\pi}{6} = \frac{1}{\sin\frac{5\pi}{6}} = \frac{1}{1/2} = 2.

Watch out

A common mistake is to forget the sign of sine in the second quadrant. Since sin⁡5π6\sin\frac{5\pi}{6} is positive, csc⁡\csc is also positive. If you mistakenly used the cosine sign (negative), you’d get a wrong result.

  1. Substitute into the expression The given expression becomes:

cot⁡2π6+csc⁡5π6+3tan⁡2π6=3+2+3⋅13\cot^2\frac{\pi}{6} + \csc\frac{5\pi}{6} + 3\tan^2\frac{\pi}{6} = 3 + 2 + 3 \cdot \frac{1}{3}

  1. Simplify

3+2+1=63 + 2 + 1 = 6

Thus the left-hand side equals 6, which is exactly the right-hand side. The identity is proved.

Tip

Notice that cot⁡2π6=3\cot^2\frac{\pi}{6} = 3 and 3tan⁡2π6=13\tan^2\frac{\pi}{6} = 1 — these two terms together give 4, and adding csc⁡5π6=2\csc\frac{5\pi}{6} = 2 completes the sum to 6. This pattern is a quick mental check.

✓Final answer

The value of the expression is 6\boxed{6}, proving the given equality.

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