Show that- 4sin-6sin 2-3 - 3cos-3tan-4 - cosec 2-2 - 2 2-4

Substituting the value of sinπ6=12, sinπ3=√(3)2, cosπ3=12, tanπ4=1 and π2=1 we have 4sinπ6sin^2π3+3cosπ3tanπ4+^2π2 =4×12×34+3×12× 1+1 ...

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Substitution Method to Remove Indeterminate Form

Show that: 4...

Question

Show that:
$4 sin \frac{π}{6} {sin}^{2} \frac{π}{3} + 3 cos \frac{π}{3} tan \frac{π}{4} + {cosec}^{2} \frac{π}{2} = 2 {sec}^{2} \frac{π}{4}$

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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}

x sin \frac{π}{6} {cos}^{2} \frac{π}{4} = \frac{{cot}^{2} \frac{π}{6} sec \frac{π}{3} tan \frac{π}{4}}{c o s e c^{2} \frac{π}{4} c o s e c \frac{π}{6}}

x sin \frac{π}{6} {cos}^{2} \frac{π}{4} = \frac{{cot}^{2} \frac{π}{6} sec \frac{π}{3} tan \frac{π}{4}}{c o s e c^{2} \frac{π}{4} c o s e c \frac{π}{6}}

A + B + C = π

sin (\frac{A}{2}) + sin (\frac{B}{2}) + sin (\frac{C}{2})

1 + 4 sin \frac{π - A}{4} sin \frac{π - B}{4} sin \frac{π - C}{4}

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

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