Question

5.Prove that :
sin 19° + cos 11°÷
cos 19°- sin 11°=√3​

Answer 1

Given : sin 19° + cos 11°÷  cos 19°- sin 11°=√3​

To Find :  Prove

Solution:

Use Sin(90° - x) = Cosx

Cos(90° - x) = Sinx

SinA + SinB = 2Sin((A + B)/2) Cos((A- B)/2)

SinA - SinB = 2Sin((A - B)/2) Cos((A+ B)/2)

(sin 19° + cos 11°)/(cos 19°- sin 11°)

Numerator :

sin 19° + cos 11°

= Sin19° + Sin79°

= 2Sin((19°+ 79°)/2)(Cos(19° - 79°)/2)

= 2Sin(49°)Cos(-30°)

=2 Sin(49°)Cos(30°)

= 2Sin(49°)(√3/2)

=√3 Sin(49°)

Denominator

cos 19°- sin 11°

=Sin71° - Sin11°

= 2Sin((71°  - 11°)/2)Cos((71° + 11°)/2)

= 2Sin30°Cos41°

= 2(1/2)Cos41°

= Cos41°

= Sin49°

LHS  =

Numerator/ Denominator

= √3 Sin(49°) / Sin49°

=√3

= RHS

QED

Hence Proved

sin 19° + cos 11°÷  cos 19°- sin 11°=√3​  

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Answer 2

SOLUTION

TO PROVE

$\displaystyle \sf{ \frac{ \sin {19}^{ \circ} + \cos {11}^{ \circ} }{ \cos {19}^{ \circ} - \sin {11}^{ \circ} } = \sqrt{3} }$

PROOF

LHS

$= \displaystyle \sf{ \frac{ \sin {19}^{ \circ} + \cos {11}^{ \circ} }{ \cos {19}^{ \circ} - \sin {11}^{ \circ} } }$

$= \displaystyle \sf{ \frac{ \sin {19}^{ \circ} + \cos ( {90}^{ \circ} - {79}^{ \circ}) }{ \cos {19}^{ \circ} - \sin ( {90}^{ \circ} - {79}^{ \circ}) } }$

$= \displaystyle \sf{ \frac{ \sin {19}^{ \circ} + \sin {79}^{ \circ} }{ \cos {19}^{ \circ} - \cos {79}^{ \circ} } }$

$= \displaystyle \sf{ \frac{ 2\sin \big( \frac{{19}^{ \circ} + {79}^{ \circ} }{2} \big) \cos \big( \frac{{19}^{ \circ} - {79}^{ \circ} }{2} \big) }{ 2\sin \big( \frac{{19}^{ \circ} + {79}^{ \circ} }{2} \big) \sin \big( \frac{{19}^{ \circ} - {79}^{ \circ} }{2} \big) } }$

$= \displaystyle \sf{ \frac{ 2\sin {49}^{ \circ} \cos {30}^{ \circ} }{ 2\sin {49}^{ \circ} \sin {30}^{ \circ} } }$

$= \displaystyle \sf{ \frac{\cos {30}^{ \circ} }{ \sin {30}^{ \circ} } }$

= cot 30°

$= \sqrt{3}$

= RHS

Hence proved

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