Question 19 of 50beginner🔧 ApplyShort Answer2 marks

Prove that: 2sin² π/6 + cosec² 7π/6 cos² π/3 = 3/2

Correct Answer

L.H.S. = 2sin² π/6 + cosec² 7π/6 cos² π/3

We know that sin π/6 = 1/2 and cos π/3 = 1/2. Therefore, 2sin² π/6 = 2 × (1/2)² = 2×12 \times 1/4 = 1/2

Also, 7π/6 = π + π/6, which lies in the third quadrant where sine is negative.

So, sin 7π/6 = sin(π + π/6) = -sin π/6 = -1/2

Therefore, cosec 7π/6 = 1/sin 7π/6 = 1/(-1/2) = -2

And cosec² 7π/6 = (-2)² = 4

Now, cosec² 7π/6 × cos² π/3 = 4 × (1/2)² = 4×14 \times 1/4 = 1

L.H.S. = 1/2 + 1 = 3/2 = R.H.S.

Hence proved.

Exercise: EXERCISE 3.3 | Q: 2 | (Page 25)
For More Understanding

Explanation

This question from Exercise 3.3 tests students on evaluating trigonometric functions at standard angles and applying quadrant rules. The key steps involve: (1) substituting standard values sin π/6 = 1/2 and cos π/3 = 1/2, (2) recognizing that 7π/6 is in the third quadrant where sine is negative, and (3) using the identity sin(π + θ) = -sin θ. The calculation follows the same pattern as Example 10 which demonstrates evaluating expressions with angles like 5π/6 = π - π/6.

Solution Steps

  1. Step 1: Evaluate 2sin² π/6 = 2 × (1/2)² = 1/2

  2. Step 2: Find sin 7π/6 = sin(π + π/6) = -sin π/6 = -1/2

  3. Step 3: Calculate cosec² 7π/6 = (-2)² = 4

  4. Step 4: Evaluate cosec² 7π/6 × cos² π/3 = 4×14 \times 1/4 = 1

  5. Step 5: Add results: 1/2 + 1 = 3/2 = R.H.S.

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