Introduction to...

$$\dfrac { \tan ^{ 2 }{ 60^0 } +4\sin ^{ 2 }{ 45^0 } +3\sec ^{ 2 }{ 30 ^0} +5\cos ^{ 2 }{ 90^0 }  }{ \csc { 30^0 } +\sec { 60 ^0} -\cot ^{ 2 }{ 30^0 }  }  =$$

If $$(\sin \theta - \cos \theta) = 0$$, then $$(\sin^{4} \theta + \cos^{4} \theta) =$$

$$\dfrac { 1 }{ 4 } \left[ \cot ^{ 4 }{ 30^0 } -\csc ^{ 4 }{ 60^0 }  \right] +\dfrac { 3 }{ 2 } \left[ \sec ^{ 2 }{ 45^0 } -\tan ^{ 2 }{ 30^0 }  \right] -5\cos ^{ 2 }{ 60^0 } =$$

$$4\left( \sin ^{ 4 }{ 30^0} +\cos ^{ 4 }{ 60^0 }  \right) -\dfrac { 2 }{ 3 } \left( \sin ^{ 2 }{ 60^0 } -\cos ^{ 2 }{ 45^0 }  \right) +\dfrac { 1 }{ 2 } \tan ^{ 2 }{ 60^0 } =$$

$$2\sin 30˚ + 2\tan 45˚ - 3\cos 60˚ - 2\cos^{2} 30˚ =$$

$$\sin 30^{\circ} \cos 60^{\circ} + \sin 60^{\circ} \cos 30^{\circ}  $$ is equal to

If $$\cos ^{ 2 }{ 45^o- } \cos ^{ 2 }{ 30^o=x\cos {  { 45^o\sin {  { 45^o }  }  }  }  } $$, then $$x = $$

$$\cos^{2}45^0 + \sin^{2}60^0 + \sin^{2} 30^0 = $$

Solve: $$\sec 70^{\circ} \sin 20^{\circ} + \cos 20^{\circ} \text{cosec } 70^{\circ} $$

$$\sec (90 - \theta) = $$