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प्रश्न
Prove that:
cot12° cot38° cot52° cot60° cot78° = \[\frac{1}{\sqrt{3}}\]
बेरीज
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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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