If A - B - C - prove thatA-2-B-2-C-2-A-2B-2C-2

Here, A + B + C = π ⇒ ( A/2 + B/2 )= ( π/2 C/2 ) ⇒cot ( A/2+B/2 ) = cot ( π/2 C/2 )= tanC/2 ⇒cotA/2cotB/2 1/cotA/2 + cotB/2 = 1/cotC/2 ⇒cotA/2+cotB/2 = cotA/ ...

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Byju's Answer

Standard XII

Mathematics

Composition of Trigonometric Functions and Inverse Trigonometric Functions

If A + B + C ...

Question

If A + B + C = $π$ , prove that

$cot \frac{A}{2} + cot \frac{B}{2} + cot \frac{C}{2} = cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2}$

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Solution

Here, A + B + C = $π$

$\Rightarrow (\frac{A}{2} + \frac{B}{2}) = (\frac{π}{2} - \frac{C}{2})$

$\Rightarrow cot (\frac{A}{2} + \frac{B}{2}) = cot (\frac{π}{2} - \frac{C}{2}) = tan \frac{C}{2}$

$\Rightarrow \frac{cot \frac{A}{2} cot \frac{B}{2} - 1}{cot \frac{A}{2} + cot \frac{B}{2}} = \frac{1}{cot \frac{C}{2}}$

$\Rightarrow cot \frac{A}{2} + cot \frac{B}{2} = cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2} - cot \frac{C}{2}$

$\Rightarrow cot + cot \frac{B}{2} + cot \frac{C}{2} = cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2}$

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If A + B + C = π, prove that cot A2 + cot B2+cot C2=cot A2 cot B2cot C2

Here, A + B + C = π ⇒(A2+B2)=(π2−C2) ⇒cot(A2+B2)=cot(π2−C2)=tanC2 ⇒cotA2cotB2−1cotA2+cotB2=1cotC2 ⇒cotA2+cotB2=cotA2cotB2cotC2−cotC2 ⇒cot+cotB2+cotC2=cotA2cotB2cotC2

If A + B + C = $π$ , prove that

$cot \frac{A}{2} + cot \frac{B}{2} + cot \frac{C}{2} = cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2}$