$sin (45^{o} + A) - cos (45^{o} - A)$ is equal to .

\frac{1}{\sqrt{2}}

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1

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\sqrt{2}

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Solution

The correct option is C $0$
$s i n (45^{\circ} + A) - c o s (45^{\circ} - A)$
$\Rightarrow s i n 45^{\circ} c o s A + c o s 45^{\circ} s i n A - [c o s 75^{\circ} c o s A + s i n 75^{\circ} s i n A]$
$\Rightarrow \frac{1}{\sqrt{2}} c o s A + \frac{1}{\sqrt{2}} s i n A - \frac{1}{\sqrt{2}} c o s A - s i n 45^{\circ} s i n A$
$\Rightarrow 0$
$[∵ s i n (A + B) = s i n A c o s B + c o s A s i n B$
$c o s (A - B) = c o s A c o s B + s i n A s i n B]$
Option C is correct

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Similar questions

\frac{sin 30^{o} + cos 45^{o} - tan 60^{o}}{cot 30^{o} - sin 45^{o} - cos 60^{o}} = - 1

Evaluate the following.
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sin 45^{o} + cos 45^{o}

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cos 30^{o} = sin 60^{o} = \frac{\sqrt{3}}{2}

cos 60^{o} = sin 30^{o} = \frac{1}{2}

cos 45^{o} = sin 45^{o} = \frac{1}{\sqrt{2}}

\int_{0}^{1} \frac{2 {sin}^{- 1} \frac{x}{2}}{x} d x

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