Question

solve 15th one.
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Answer 1

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Answer 2

SOLUTION

TO PROVE

$\displaystyle \sf{ \frac{ \sin {20}^{ \circ} }{ \cos {70}^{ \circ}} + \frac{ \tan {70}^{ \circ}}{ \cot {20}^{ \circ}} - 2 \sec {20}^{ \circ} \sin {70}^{ \circ} = 0}$

FORMULA TO BE IMPLEMENTED

$\sf{1. \: \: cos( {90}^{ \circ} - \theta) = \sin \theta}$

$\sf{2. \: \: cot( {90}^{ \circ} - \theta) = \tan \theta}$

$\sf{3. \: \: sin( {90}^{ \circ} - \theta) = \cos \theta}$

PROOF

LHS

$\displaystyle \sf{ = \frac{ \sin {20}^{ \circ} }{ \cos {70}^{ \circ}} + \frac{ \tan {70}^{ \circ}}{ \cot {20}^{ \circ}} - 2 \sec {20}^{ \circ} \sin {70}^{ \circ} }$

$\displaystyle \sf{ = \frac{ \sin {20}^{ \circ} }{ \cos ( {90}^{ \circ} - {20}^{ \circ})} + \frac{ \tan {70}^{ \circ}}{ \cot ( {90}^{ \circ} - {70}^{ \circ})} - 2 \sec {20}^{ \circ} \sin ( {90}^{ \circ} - {20}^{ \circ} )}$

$\displaystyle \sf{ = \frac{ \sin {20}^{ \circ} }{ \sin {20}^{ \circ}} + \frac{ \tan {70}^{ \circ}}{ \tan {70}^{ \circ}} - 2 \sec {20}^{ \circ} \cos {20}^{ \circ} }$

$\displaystyle \sf{ = 1 + 1 - 2 \times \frac{1}{\cos {20}^{ \circ}} \times \cos {20}^{ \circ} }$

$= 2 - 2$

$= 0$

= RHS

Hence proved

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