Ifseco + Coso = 2Prove Sec100o+cos100o = 2

1.	Ifseco + Coso = 2Prove Sec100o+cos100o = 2
Answer» Let's take the angle as ALPHA for clarity purposes. $= > \sec( \alpha ) + \cos( \alpha ) = 2 \\ = > \frac{1}{ \cos( \alpha ) } + \cos( \alpha ) = 2 \\ = > \frac{1 + \cos^{2} ( \alpha ) }{ \cos( \alpha ) } = 2 \\ = > \cos^{2} ( \alpha ) - 2 \cos( \alpha ) + 1 = 0 \\ = > ( \cos( \alpha ) - 1)^{2} = 0 \\ = > \cos( \alpha ) = 1 \\ = > \alpha = 2n\pi, \: \: where \: n \: is \: an \: integer \\$ If we take the PRINCIPAL value of cosine then, $\alpha = 0$ Now, $\sec(100 \alpha ) + \cos(100 \alpha ) = \sec(0) + \cos(0) = 1 + 1 = 2$ *- Q.E.D.* Hope, it'll HELP you.....

Answer»

Let's take the angle as ALPHA for clarity purposes.

$= > \sec( \alpha ) + \cos( \alpha ) = 2 \\ = > \frac{1}{ \cos( \alpha ) } + \cos( \alpha ) = 2 \\ = > \frac{1 + \cos^{2} ( \alpha ) }{ \cos( \alpha ) } = 2 \\ = > \cos^{2} ( \alpha ) - 2 \cos( \alpha ) + 1 = 0 \\ = > ( \cos( \alpha ) - 1)^{2} = 0 \\ = > \cos( \alpha ) = 1 \\ = > \alpha = 2n\pi, \: \: where \: n \: is \: an \: integer \\$