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Question: Answered & Verified by Expert
If $\mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{\pi}{2}$, then
$$
\sqrt{2} \cos \left(\frac{\pi}{4}-A\right)+\sqrt{2} \cos \left(\frac{\pi}{4}-B\right)+\sqrt{2} \cos \left(\frac{\pi}{4}-C\right)+1=
$$
MathematicsTrigonometric Ratios & IdentitiesTS EAMCETTS EAMCET 2022 (20 Jul Shift 1)
Options:
  • A $4 \sqrt{2} \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
  • B $4 \cos \frac{\mathrm{A}}{2} \cos \frac{\mathrm{B}}{2} \cos \frac{\mathrm{C}}{2}$
  • C $4 \sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$
  • D $4 \sqrt{2} \sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$
Solution:
1794 Upvotes Verified Answer
The correct answer is: $4 \sqrt{2} \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
$\begin{aligned} & \sqrt{2} \cos \left(\frac{\pi}{4}-\mathrm{A}\right)+\sqrt{2} \cos \left(\frac{\pi}{4}-\mathrm{B}\right)+ \\ & \qquad \sqrt{2} \cos \left(\frac{\pi}{4}-\mathrm{C}\right)+\sqrt{2} \cos \left(\frac{\pi}{4}\right)\end{aligned}$
$\begin{aligned}
& \text { Here, } \mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{\pi}{2} \\
& =\sqrt{2}\left[\left\{\cos \left(\frac{\pi}{4}-\mathrm{A}\right)+\cos \left(\frac{\pi}{4}-\mathrm{B}\right)\right\}+\left\{\cos \left(\frac{\pi}{4}-\mathrm{C}\right)+\cos \left(\frac{\pi}{4}\right)\right\}\right] \\
& =\sqrt{2}\left[2 \cos \left(\frac{\frac{\pi}{2}-(\mathrm{A}+\mathrm{B})}{2}\right) \cos \left(\frac{\mathrm{B}-\mathrm{A}}{2}\right)+2 \cos \left(\frac{\frac{\pi}{4}-\mathrm{C}+\frac{\pi}{4}}{2}\right) \cos \frac{\left(\frac{\pi}{4}-\mathrm{C}-\frac{\pi}{4}\right)}{2}\right] \\
& =2 \sqrt{2}\left[\cos \left(\frac{\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{B}-\mathrm{A}}{2}\right)+\cos \left(\frac{\mathrm{C}}{2}\right) \cos \left(\frac{\pi}{4}-\frac{\mathrm{C}}{2}\right)\right] \\
& =2 \sqrt{2} \cos \left(\frac{\mathrm{C}}{2}\right)\left[\cos \left(\frac{\mathrm{B}-\mathrm{A}}{2}\right)+\cos \left(\frac{\pi}{4}-\frac{\mathrm{C}}{2}\right)\right] \\
& =2 \sqrt{2} \cos \left(\frac{\mathrm{C}}{2}\right)\left[2 \cos \left(\frac{\frac{\mathrm{B}-\mathrm{A}}{2}+\frac{\pi}{4}-\frac{\mathrm{C}}{2}}{2}\right) \cos \left(\frac{\frac{\mathrm{B}-\mathrm{A}}{2}-\frac{\pi}{4}+\frac{\mathrm{C}}{2}}{2}\right)\right] \\
& =4 \sqrt{2} \cos \left(\frac{\mathrm{C}}{2}\right)\left[\cos \left(\frac{\frac{\mathrm{B}}{2}-\frac{1}{2}\left(\frac{\pi}{2}-\mathrm{B}\right)+\frac{\pi}{4}}{2}\right) \cos \left(\frac{\frac{1}{2}\left(\frac{\pi}{4}-\mathrm{A}\right)-\frac{\pi}{4}-\frac{\mathrm{A}}{2}}{2}\right)\right] \\
& =4 \sqrt{2} \cos \left(\frac{\mathrm{C}}{2}\right)\left[\cos \left(\frac{\frac{\mathrm{B}}{2}-\frac{\pi}{2}+\frac{\mathrm{B}}{2}+\frac{\pi}{4}}{2}\right) \cos \left(\frac{\frac{\pi}{4}-\frac{\mathrm{A}}{2}-\frac{5}{4}-\frac{\mathrm{A}}{2}}{2}\right)\right] \\
& =4 \sqrt{2} \cos \left(\frac{\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}}{2}\right) \\
\end{aligned}$
So, option (a) is correct.

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