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NCERT Class XI Mathematics - Trigonometric Functions - Solutions

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Question : 19 of 61
Marks: +1, -0
2sin⁡2π6+csc⁡27π6cos⁡2π32\sin^2 \frac{\pi}{6} + \csc^2 \frac{7\pi}{6} \cos^2 \frac{\pi}{3} = 32\frac{3}{2}
Solution:  
L..S = 2sin⁡2π6+csc⁡27π6cos⁡2π32\sin^2 \frac{\pi}{6} + \csc^2 \frac{7\pi}{6} \cos^2 \frac{\pi}{3}
= 2 × 14\frac{1}{4} + (csc⁡2(π+π6))(12)2\left(\csc^2 \left(\pi + \frac{\pi}{6}\right)\right) \left(\frac{1}{2}\right)^2 = 12+(−2)2×14\frac{1}{2} + (-2)^2 \times \frac{1}{4} = 12+1\frac{1}{2} + 1 = 32\frac{3}{2} = R.H.S.
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