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NCERT Class XI Mathematics - Trigonometric Functions - Solutions

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Question : 21 of 61
Marks: +1, -0
2sin⁡23π4+2cos⁡2π4+2sec⁡2π32\sin^2\frac{3\pi}{4} + 2\cos^2\frac{\pi}{4} + 2\sec^2\frac{\pi}{3} = 10
Solution:  
L.H.S. = 2sin⁡23π4+2cos⁡2π4+2sec⁡2π32\sin^2\frac{3\pi}{4} + 2\cos^2\frac{\pi}{4} + 2\sec^2\frac{\pi}{3}
= 2sin⁡24π−π42\sin^2\frac{4\pi-\pi}{4} + 2 × (12)2\left(\frac{1}{\sqrt{2}}\right)^2 + 2 × (2)2(2)^2 = 2sin⁡2(π−π4)2\sin^2\left(\pi - \frac{\pi}{4}\right) + 2 × 12\frac{1}{2} + 8
= 2sin⁡2π4+1+82\sin^2\frac{\pi}{4} + 1 + 8 = 2 × (12)2\left(\frac{1}{\sqrt{2}}\right)^2 + 9 = 10 = R.H.S.
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