Question

prove cot(pi/12-A) + tan(pi/12+A)=4cos2A/1-2sin2A

Answer 1

SOLUTION

TO PROVE

$\displaystyle \sf{ \cot \bigg( \frac{\pi}{12} -A \bigg) + \tan \bigg( \frac{\pi}{12} + A \bigg) = \frac{4 \cos 2A}{1 - 2 \sin 2A} }$

FORMULA TO BE IMPLEMENTED

1. cos (A - B) = cos A cos B + sin A sin B

2. 2 sin A cos B = sin (A + B) + sin (A - B)

EVALUATION

LHS

$\displaystyle \sf{ = \cot \bigg( \frac{\pi}{12} -A \bigg) + \tan \bigg( \frac{\pi}{12} + A \bigg) }$

$\displaystyle \sf{ = \frac{\cos \bigg( \frac{\pi}{12} -A \bigg)}{\sin \bigg( \frac{\pi}{12} -A \bigg)} + \frac{\sin \bigg( \frac{\pi}{12} + A \bigg)}{\cos \bigg( \frac{\pi}{12} + A \bigg)} }$

$\displaystyle \sf{ = \frac{\cos \bigg( \frac{\pi}{12} + A \bigg)\cos \bigg( \frac{\pi}{12} -A \bigg) + \sin \bigg( \frac{\pi}{12} + A \bigg)\sin \bigg( \frac{\pi}{12} - A \bigg)}{\sin \bigg( \frac{\pi}{12} -A \bigg)\cos \bigg( \frac{\pi}{12} + A \bigg)} }$

$\displaystyle \sf{ = \frac{\cos \bigg( \frac{\pi}{12} + A - \frac{\pi}{12} + A \bigg) }{\sin \bigg( \frac{\pi}{12} -A \bigg)\cos \bigg( \frac{\pi}{12} + A \bigg)} }$

$\displaystyle \sf{ = \frac{\cos 2 A }{\sin \bigg( \frac{\pi}{12} -A \bigg)\cos \bigg( \frac{\pi}{12} + A \bigg)} }$

$\displaystyle \sf{ = \frac{2\cos 2 A }{2 \: \sin \bigg( \frac{\pi}{12} -A \bigg)\cos \bigg( \frac{\pi}{12} + A \bigg)} }$

$\displaystyle \sf{ = \frac{2\cos 2 A }{ \: \sin \bigg( \frac{\pi}{12} -A + \frac{\pi}{12} + A \bigg) + \sin \bigg( \frac{\pi}{12} -A - \frac{\pi}{12} - A \bigg)} }$

$\displaystyle \sf{ = \frac{2\cos 2 A }{ \: \sin \frac{\pi}{6} + \sin ( -2A )} }$

$\displaystyle \sf{ = \frac{2\cos 2 A }{ \: \frac{1}{2} - \sin 2A } }$

$\displaystyle \sf{ = \frac{4\cos 2 A }{ \: 1 - 2\sin 2A } }$

= RHS

Hence proved

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