Prove that-sin2 -6-cos 2-3-tan 2-4-1-2

We have L.H.S = sin^2 π/6 + cos^2 π/3 tan^2 π/4 = ( 1/2)^2 + (1/2)^2 (1)^2 = 1/4+ 1/4 1 = 1+1 4/4 = 2/4 = 1/2= R.H.S ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Sin2a and Cos2a in Terms of Tana

Prove that:si...

Question

Prove that:

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

Open in App

Solution

Suggest Corrections

Similar questions

{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}

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}

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

Prove that:

$s i n^{2} (\frac{π}{8} + \frac{A}{2}) - s i n^{2} (\frac{π}{8} - \frac{A}{2}) = \frac{1}{\sqrt{2}} s i n A$

Join BYJU'S Learning Program