MCQ
જો $\sin A + \cos A = \sqrt 2 ,$ તો ${\cos ^2}A = $
- A$\frac{1}{4}$
- ✓$\frac{1}{2}$
- C$\frac{1}{{\sqrt 2 }}$
- D$\frac{3}{2}$
On squaring both the sides
==> $1 + \sin 2A = 2\, \Rightarrow \sin 2A = 1 = \sin {90^o}$
==> $2A = {90^o}$ or $A = {45^o}$
Now, ${\cos ^2}A = {(\cos {45^o})^2} $
$= {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} = \frac{1}{2}$.
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