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

LHS = sin^2 π/6 + cos ^2 π/3 tan ^2 π/4 = ( 1/2 )^2+ ( 1/2 )^2 1^2 [ ∵ sin π/6 = 1/2, cos π/3 = 1/2 and tan π/4 = 1 ] = ( 1/4+ 1/4 1 ) = 1/2 ...

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Sin2a and Cos2a in Terms of Tana

Prove that si...

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Prove that sin $^{2} \frac{π}{6}$ + cos $^{2} \frac{π}{3}$ - tan $^{2} \frac{π}{4}$ = - $\frac{1}{2}$ .

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

sin \frac{π}{6} cos 0 + sin \frac{π}{4} cos \frac{π}{4} + sin \frac{π}{3} cos \frac{π}{6} = \frac{7}{4}

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

{sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} + {cot}^{2} \frac{π}{2}

\frac{1 - {tan}^{2} (\frac{π}{4} - A)}{1 + {tan}^{2} (\frac{π}{4} - A)} = sin b A

{cos}^{2} (\frac{π}{6} + θ) - {sin}^{2} (\frac{π}{6} - θ)

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