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3.Trigonometrical Ratios, Functions and Identities
easy
यदि $\sin A + \cos A = \sqrt 2 ,$ तो ${\cos ^2}A = $
A
$\frac{1}{4}$
B
$\frac{1}{2}$
C
$\frac{1}{{\sqrt 2 }}$
D
$\frac{3}{2}$
Solution
$\sin A + \cos A = \sqrt 2 $
दोनों तरफ वर्ग करने पर,
$\Rightarrow 1 + \sin 2A = 2\, \Rightarrow \sin 2A = 1 = \sin {90^o}$
$\Rightarrow 2A = {90^o}$या $A = {45^o}$
अब ${\cos ^2}A = {(\cos {45^o})^2}$
$= {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} = \frac{1}{2}$.
Standard 11
Mathematics
