Question

Value
of
3 + cot 80 cot 20/cot80+cot20 is equal to​

Answer 1

SOLUTION :

TO DETERMINE

The value of

$\displaystyle \sf{ \frac{3 + \cot {80}^{ \circ} \cot {20}^{ \circ} }{ \cot {80}^{ \circ} + \cot {20}^{ \circ}} \: }$

EVALUATION

$\displaystyle \sf{ \frac{3 + \cot {80}^{ \circ} \cot {20}^{ \circ} }{ \cot {80}^{ \circ} + \cot {20}^{ \circ}} \: }$

$= \displaystyle \sf{ \frac{3 + \frac{ \cos {80}^{ \circ} \cos {20}^{ \circ}}{ \sin {80}^{ \circ} \sin {20}^{ \circ}} }{ \frac{\cos {80}^{ \circ} }{\sin{80}^{ \circ} } + \frac{\cos {20}^{ \circ} }{\sin{20}^{ \circ} } } \: }$

$= \displaystyle \sf{ \frac{3 \sin {80}^{ \circ} \sin {20}^{ \circ} + \cos {80}^{ \circ} \cos {20}^{ \circ} }{\sin {80}^{ \circ} \cos {20}^{ \circ} + \cos {80}^{ \circ} \sin {20}^{ \circ}} \: }$

$= \displaystyle \sf{ \frac{3 \times 2 \sin {80}^{ \circ} \sin {20}^{ \circ} +2 \cos {80}^{ \circ} \cos {20}^{ \circ} }{2\sin ({80}^{ \circ} + {20}^{ \circ}) }}$

$= \displaystyle \sf{ \frac{3 ( \cos {60}^{ \circ} - \cos {100}^{ \circ}) + ( \cos {60}^{ \circ} + \cos {100}^{ \circ}) }{2\sin {100}^{ \circ} }}$

$= \displaystyle \sf{ \frac{4\cos {60}^{ \circ} - 2\cos {100}^{ \circ} }{2\sin {100}^{ \circ} }}$

$= \displaystyle \sf{ \frac{2\cos {60}^{ \circ} - \cos {100}^{ \circ} }{\sin {100}^{ \circ} }}$

$= \displaystyle \sf{ \frac{2 \times \frac{1}{2} - \cos {100}^{ \circ} }{\sin {100}^{ \circ} }}$

$= \displaystyle \sf{ \frac{1 - \cos {100}^{ \circ} }{\sin {100}^{ \circ} }}$

$= \displaystyle \sf{ \frac{2 {\sin}^{2} {50}^{ \circ} }{2\sin {50}^{ \circ}\cos {50}^{ \circ} }}$

$= \displaystyle \sf{ \frac{ {\sin} {50}^{ \circ} }{\cos {50}^{ \circ} }}$

$= \sf{\tan {50}^{ \circ}}$

RESULT

$\boxed{ \displaystyle \sf{ \: \: \frac{3 + \cot {80}^{ \circ} \cot {20}^{ \circ} }{ \cot {80}^{ \circ} + \cot {20}^{ \circ}} \: } = \tan {50}^{ \circ} \: \: \: \: \: }$

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(b+c-a)(cotB/2+cotC/2)=2a×cotA/2

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