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S I N θ Cos ( 90 ° − θ ) C O S θ Sin ( 90 ° − θ ) + C O S θ Sin ( 90 ° − θ ) S I N θ Cos ( 90 ° − θ ) - Mathematics

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प्रश्न

Prove that:

\[\frac{sin\theta  \cos(90° - \theta)cos\theta}{\sin(90° - \theta)} + \frac{cos\theta  \sin(90° - \theta)sin\theta}{\cos(90° - \theta)}\]

योग

उत्तर

\[\ LHS = \frac{\sec\left( 90° - \theta \right) cosec\theta - \tan\left( 90° - \theta \right) \cot\theta + \cos^2 25° + \cos^2 65°}{3\tan27° \tan63°}\]

\[ = \frac{cosec\theta cosec\theta - \cot\theta \cot\theta + \sin^2 \left( 90° - 25° \right) + \cos^2 65°}{3\tan27° \cot\left( 90° - 63° \right)}\]

\[ = \frac{{cosec}^2 \theta - \cot^2 \theta + \sin^2 65° + \cos^2 65°}{3\tan27°\cot27°}\]

\[ = \frac{1 + 1}{3 \times \tan27° \times \frac{1}{\tan27°}}\]

\[ = \frac{2}{3}\]

= RHS

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अध्याय 7: Trigonometric Ratios of Complementary Angles - Exercises [पृष्ठ ३१३]

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आरएस अग्रवाल Mathematics [English] Class 10
अध्याय 7 Trigonometric Ratios of Complementary Angles
Exercises | Q 5.6 | पृष्ठ ३१३

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