Plzz ans this as quick as possible

1.	Plzz ans this as quick as possible
Answer» we know that ${\sin} {}^{2} ( \alpha ) + { \cos} {}^{2} ( \alpha ) = 1$ so ${\cos} {}^{2} ( \alpha ) = 1 - { \sin} {}^{2} ( \alpha )$ $but \: we \: have \: sin \alpha = \sin(q) \\ more \: appropritel \sin( \alpha ) = a \div b$ THEREFORE ${ \sin(q) } {}^{2} = (a \div b) {}^{2}$ $\cos( = \alpha ) = \cos(q)$ ${\cos(q) } {}^{2} = 1 - a {}^{2} \div b {}^{2}$ $we \: have \: \\ \cos(q) = \sqrt{b {}^{2} - a {}^{2} \div b {}^{2} }$ therefore OPTION C should have been CORRECT but it is having a little blunder $i.e. \: b {}^{2} must \: also \: be \: placed \: under \: the \: root \:$

Answer»

we know that

${\sin} {}^{2} ( \alpha ) + { \cos} {}^{2} ( \alpha ) = 1$