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प्रश्न
If x lies in the first quadrant and \[\cos x = \frac{8}{17}\], then prove that:
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उत्तर
\[\text{ Given: }0 < x < \frac{\pi}{2}\]
\[\text{ Now, }\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \frac{64}{289}} = \frac{15}{17}\]
\[\text{ LHS }= \cos\left( \frac{\pi}{6} + x \right) + \cos\left( \frac{\pi}{4} - x \right) + \cos\left( \frac{2\pi}{3} - x \right)\]
\[ = \cos(30 + x) + \cos(45 - x) + \cos(120 - x)\]
\[ = \cos 30^\circ \cos x - \sin30^\circ \sin x + \cos 45^\circ \cos x + \sin 45^\circ \sin x + \cos120^\circ \cos x + \sin120^\circ \sin x \left\{\text{ Using formulas of }\cos(A + B)\text{ and }\cos(A - B \right\})\]
\[ = \cos x(\cos 30^\circ + \cos 45^\circ + \cos120) + \sin x( - \sin 30^\circ + \sin 45^\circ + \sin 120^\circ)\]
\[ = \frac{8}{17}\left( \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} - \frac{1}{2} \right) + \frac{15}{17}\left( - \frac{1}{2} + \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \right) \]
\[ = \frac{8}{17}\left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} \right) + \frac{15}{17}\left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} \right)\]
\[ = \frac{23}{17}\left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} \right) \]
= RHS
Hence proved .
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