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In a ∆Abc, Prove That: Tan a + B 2 = Cot C 2 - Mathematics

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प्रश्न

In a ∆ABC, prove that:

\[\tan\frac{A + B}{2} = \cot\frac{C}{2}\]
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उत्तर

 In ∆ ABC: 
\[A + B + C = \pi\]
\[ \Rightarrow A + B = \pi - C\]
\[ \Rightarrow \frac{A + B}{2} = \frac{\pi - C}{2}\]
\[ \Rightarrow \frac{A + B}{2} = \frac{\pi}{2} - \frac{C}{2}\]
\[\text{ Now, LHS }= \tan\left( \frac{A + B}{2} \right) \]
\[ = \tan\left( \frac{\pi}{2} - \frac{C}{2} \right) \]
\[ = \cot\left( \frac{C}{2} \right) \left[ \because \tan\left( \frac{\pi}{2} - \theta \right) = \cot \theta \right] \]
 = RHS
Hence proved. 

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अध्याय 5: Trigonometric Functions - Exercise 5.3 [पृष्ठ ४०]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 5 Trigonometric Functions
Exercise 5.3 | Q 6.3 | पृष्ठ ४०

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