Prove that tan 11-3 - 2sin4-6 - 3-4cos ec2-4 - 4cos 217-6 - 3 - 4-3-2

LHS =tan11π3 2sin4π3 34^2π4+ 4cos^217π6 π4=√(2) =tan(3π+2π3) 2sin2π3 34(√(2))^2+ 4cos^2(2π+5π6) =tan2π3 ...

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Addition Rule of Differentiation

Prove that ...

Question

Prove that $tan \frac{11 π}{3} - 2 sin \frac{4 π}{6} - \frac{3}{4} cos e c^{2} \frac{π}{4} + 4 {cos}^{2} \frac{17 π}{6} = \frac{3 - 4 \sqrt{3}}{2}$

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tan \frac{11 π}{3} - 2 sin \frac{4 π}{6} - \frac{3}{4} {csc}^{2} \frac{π}{4} + 4 {cos}^{2} \frac{17 π}{6} = \frac{3 - 4 \sqrt{3}}{2}

Prove that:

$(i) t a n 225^{\circ} c o t 405^{\circ} + t a n 765^{\circ} c o t 675^{\circ} = 0$

$(i i) s i n \frac{8 π}{3} c o s \frac{23 π}{6} + c o s \frac{13 π}{3} s i n \frac{35 π}{6} = \frac{1}{2}$

$(i i i) c o s 24^{\circ} + c o s 55^{\circ} + c o s 125^{\circ} + c o s 204^{\circ} + c o s 300^{\circ} = \frac{1}{2}$

$(i v) t a n (- 225^{\circ}) c o t (- 405^{\circ}) - t a n (- 765^{\circ}) c o t (675^{\circ}) = 0$

$(v) c o s 570^{\circ} s i n 510^{\circ} + s i n (- 330^{\circ}) c o s (- 390^{\circ}) = 0$

$(v i) t a n \frac{11 π}{3} - 2 s i n \frac{4 π}{6} - \frac{3}{4} c o s e c^{2} \frac{π}{4} + 4 c o s^{2} \frac{17 π}{6} = \frac{3 - 4 \sqrt{3}}{2}$

$(v i i) 3 s i n \frac{π}{6} s e c \frac{π}{3} - 4 s i n \frac{5 π}{5} c o t \frac{π}{4} = 1$

c o t 7 \frac{1^{\circ}}{2} = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}

cot (7 {\frac{1}{2}}^{o}) = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}

A \times (B \cup C)

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