Q62 / 160 · AP EAMCET 2024 (20 May Shift 1) · Mathematics

$\frac{\sin 1^{\circ}+\sin 2^{\circ}+\ldots+\sin 89^{\circ}}{2\left(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}\right)+1}=$

  • A
    $\sqrt{2}$
  • Correct answer
    B
    $\frac{1}{\sqrt{2}}$
  • C
    $2$
  • D
    $\frac{1}{2}$

Solution

$\frac{\sin 1^{\circ}+\sin 2^{\circ}+\ldots+\sin 89^{\circ}}{2\left(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}\right)+1}$ $=\frac{\left[\begin{array}{c}\left(\sin 1^{\circ}+\sin 89^{\circ}\right)+\left(\sin 2^{\circ}+\sin 88^{\circ}\right)+\ldots+ \\ \left(\sin 44^{\circ}+\sin 46^{\circ}\right)+\sin 45^{\circ}\end{array}\right]}{2\left(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}\right)+1}$ $=\frac{\sin 45^{\circ}\left[2\left(\cos 44^{\circ}+\cos 43^{\circ}+\ldots+\cos 1^{\circ}\right)+1\right]}{2\left(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}\right)+1}$ $=\sin 45^{\circ}=\frac{1}{\sqrt{2}}$

Hence, the correct answer is option: 'B'.

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