मराठी

Prove That: C O T θ Tan ( 90 ° − θ ) − Sec ( 90 ° − θ ) C O S E C θ + √ 3 Tan 12 ° Tan 60 ° Tan 78 ° = 2 - Mathematics

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प्रश्न

Prove that:

\[cot\theta \tan\left( 90° - \theta \right) - \sec\left( 90° - \theta \right)cosec\theta + \sqrt{3}\tan12° \tan60° \tan78° = 2\]

बेरीज

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

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आर एस अग्रवाल Mathematics [English] Class 10
पाठ 7 Trigonometric Ratios of Complementary Angles
Exercises | Q 5.7 | पृष्ठ ३१३

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