Question 118

$$\sin^2 60^\circ + \cos^2 225^\circ + \tan^2 210^\circ + \cot^2 240^\circ =$$

Solution

Sin 60 = (root 3)/2
Cos 225 = cos (180 + 45) = - cos 45 = -1/(root 2)
Tan 210 = tan (180 + 30) = tan 30 = 1/(root 3)
Cot 240 = cot (180 + 60) = cot 60 = 1/(root 3)

Squaring up all terms and adding, we get answer as 23/12

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AP ICET 2nd May 2018 Shift-2

$$\sin^2 60^\circ + \cos^2 225^\circ + \tan^2 210^\circ + \cot^2 240^\circ =$$

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