Prove tan 2 - 6 c -sin 2 - 6 c -cos 2 - 3 c 2 - 4 c -cos 2 - c - 1 - 3 - 6 c - 1 3 cos - 3 c

L.H.S = tan^2( π6) + sin^2 ( π6) + cos^2( π3)^2π4 cos^2π = ( 1√(3))^2 + ( 12)^2 + ( 12)^2( √(2))^2 ( 1)^2 = 13 + 24 = 2 + 36 = 5 ...

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Derivative of Standard Functions

Prove tan 2...

Question

Prove $\frac{{tan}^{2} {(\frac{π}{6})}^{c} + {sin}^{2} {(\frac{π}{6})}^{c} + {cos}^{2} {(\frac{π}{3})}^{c}}{{sec}^{2} {(\frac{π}{4})}^{c} - {cos}^{2} π^{c}} = \frac{1}{\sqrt{3}} sec {(\frac{π}{6})}^{c} + \frac{1}{3} cos {(\frac{π}{3})}^{c}$

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{sin}^{2} 0 + {sin}^{2} {(\frac{π}{6})}^{c} + {sin}^{2} {(\frac{π}{3})}^{c} + {sin}^{2} {(\frac{π}{2})}^{c}

{sin}^{2} (\frac{π}{6}) + {cos}^{2} (\frac{π}{3}) - {tan}^{2} (\frac{π}{4}) = - \frac{1}{2}

{sin}^{2} \frac{π}{4} + {cos}^{2} \frac{π}{4} - {tan}^{2} \frac{π}{3}

| z | = 1

arg (\sqrt{\frac{(1 + z)}{(1 - z)}})

T h e v a l u e o f {cos}^{2} \frac{π}{4} + {cos}^{2} \frac{π}{6} + {sin}^{2} \frac{π}{6} + {sin}^{2} \frac{π}{3}

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