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NCERT Class XI Mathematics - Trigonometric Functions - Solutions

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Question : 18 of 61
Marks: +1, -0
sin⁡2π6+cos⁡2π3−tan⁡2π4\sin^2\frac{\pi}{6}+\cos^2\frac{\pi}{3}-\tan^2\frac{\pi}{4} = - 12\frac{1}{2}
Solution:  
L.H.S = sin⁡2π6+cos⁡2π3−tan⁡2π4\sin^2\frac{\pi}{6}+\cos^2\frac{\pi}{3}-\tan^2\frac{\pi}{4}
= [(12)2+(12)2−(1)2]\left[\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2-(1)^2\right] = 14+14\frac{1}{4}+\frac{1}{4} - 1 = −12-\frac{1}{2} = R.H.S
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