$\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)$ का मूल्य होगा

$\sqrt{2} \sin x$

$\sqrt{2} \cos x$

$\sqrt{2} \operatorname{cosec} x$

$\sqrt{2} \tan x$

Answer : Option B

Explanation :

\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)

=2 \cos \left(\frac{\frac{\pi}{4}+x+\frac{\pi}{4}-x}{2}\right) \cdot \cos \left(\frac{\frac{\pi}{4}+x-\frac{\pi}{4}+x}{2}\right)

\because \cos \mathrm{C}+\cos \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \cdot \cos \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)

=2 \cos \left(\frac{\pi}{4}\right) \cdot \cos x=\frac{2}{\sqrt{2}} \cdot \cos x=\sqrt{2} \cos x

विकल्प:

x . लगाना = 0°

$\Rightarrow \cos \frac{\pi}{4}+\cos \frac{\pi}{4}$

$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$

विकल्प (2) से

$\Rightarrow \sqrt{2} \cos 0^{\circ}$

$\pm=\sqrt{2} \times 1$

अत: विकल्प (2) उत्तर होगा।

cos⁡(π4+x)+cos⁡(π4−x)\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)cos(4π​+x)+cos(4π​−x) का मूल्य होगा