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प्रश्न
Show that sin 100° − sin 10° is positive.
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उत्तर
\[\text{ Let } f\left( \theta \right) = \sin100° - \sin10°\]
\[\text{ Multiplying and dividing by } \sqrt{1^2 + 1^2}, i . e . \text{ by } \sqrt{2}, \text{ we get }: \]
\[ \sqrt{2}\left( \frac{1}{\sqrt{2}}\sin100° - \frac{1}{\sqrt{2}}\sin10°\right)\]
\[ = \sqrt{2}\left( \cos45°\sin(90°+ 10°) - \sin45°\sin10°\right)\]
\[ = \sqrt{2}\left( \cos45°\cos10° - \sin45°\sin10° \right)\]
\[ = \sqrt{2}\cos(45°+ 10°) = \sqrt{2}co s55° , \text{ which is positive since \cos is positive in the first quadrant } .\]
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