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TG EAMCET 2 May 2025 Shift 2 Paper

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Question : 23 of 160

Marks: +1, -0

\\;\frac{ \sin\\; 1^{\circ}+ \sin\\; 2^{\circ}+. . . \sin\\; 89^{\circ} }{2 (\cos 1^{\circ}+\cos 2^{\circ}+. . .+\cos 44^{\circ}) +1}=

2
$\\;\frac{1}{\sqrt{2}}$
$\\;\frac{1}{2}$
$\sqrt{2}$

Solution:

We have,

\\;\\;\frac{ \sin\\; 1^{\circ}+ \sin\\; 2^{\circ}+. . .+ \sin\\; 89^{\circ} }{2 (\cos 1^{\circ}+\cos 2^{\circ}+. . .+\cos 44^{\circ}) +1}

\\;=\\;\frac{ ( \sin\\; 1^{\circ}+ \sin\\; 89^{\circ}) + ( \sin\\; 2^{\circ}+ \sin\\; 88^{\circ}) +. . .+ ( \sin\\; 44^{\circ}+ \sin\\; 46^{\circ}) + \sin\\; 45^{\circ} }{2 (\cos 1^{\circ}+\cos 2^{\circ}+. . .+\cos 44^{\circ}) +1}

\\;=\\;\frac{2 \sin\\; \frac{90^{\circ}}{2} \cos \frac{88^{\circ}}{2}+2 \sin\\; \frac{90^{\circ}}{2} \cos \frac{86^{\circ}}{2}+. . .+2 \sin\\; \frac{90^{\circ}}{2} \cos \frac{2^{\circ}}{2}+\frac{1}{\sqrt{2}}}{2 (\cos 1^{\circ}+\cos 2^{\circ}+. . .+\cos 44^{\circ}) +1}

\\;=\\;\frac{ \sqrt{2} (\cos 44^{\circ}+\cos 42^{\circ}+. . .+\cos 2^{\circ}+\cos 1^{\circ}) + \frac{1}{\sqrt{2}} }{2 (\cos 1^{\circ}+\cos 2^{\circ}+. . .+\cos 44^{\circ}) +1}

Let

x=\cos 1^{\circ}+\cos 2^{\circ}+. . .+\cos 44^{\circ}

=\\;\frac{ \sqrt{2} x+\frac{1}{\sqrt{2}} }{2x+1}=\\;\frac{ \frac{2x+1}{\sqrt{2}} }{2x+1}=\\;\frac{1}{\sqrt{2}}

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