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Prove That: Cot12° Cot38° Cot52° Cot60° Cot78° = 1 √ 3 - Mathematics

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प्रश्न

Prove that:

cot12° cot38° cot52° cot60° cot78° = \[\frac{1}{\sqrt{3}}\]

बेरीज

उत्तर

\[\left( ii \right) LHS = \cot12° \cot38°  \cot52 \cot60°\cot78° \]

\[ = \tan\left( 90° - 12°  \right) \times \tan\left( 90° - 38\° \right) \times \cot52° \times \frac{1}{\sqrt{3}} \times \cot78° \] 

\[ = \frac{1}{\sqrt{3}} \times \tan78° \times \tan52° \times \cot52° \times \cot78° \]

\[ = \frac{1}{\sqrt{3}} \times \tan78° \times \tan52° \times \frac{1}{\tan52t° } \times \frac{1}{\tan78° }\]

\[ = \frac{1}{\sqrt{3}}\]

= 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 6.2 | पृष्ठ ३१३

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