Question

Prove that

underoot3 cosec 20° – sec 20º = 4.​

Answer 1

$\mathfrak{\huge{\red{\underline{\underline{QUESTION}}}}}$

Prove that $\sqrt{3} \: cosec \: {20}^{\circ} - sec \: 20^{\circ} = 4$

$\mathfrak{\huge{\red{\underline{\underline {SOLUTION}}}}}$

Solving LHS,

$\implies \dfrac{\sqrt{3}}{sin \: {20}^{\circ}}- \dfrac{1}{cos \: 20^{\circ}}$

$\implies \dfrac{\sqrt{3}\:cos\: 20^{\circ} - sin \: 20^{\circ}}{sin\: 20^{\circ}. cos\: 20^{\circ}}$

Dividing numerator and denominator by 2

$\implies \dfrac {\dfrac{\sqrt{3}}{2} \: cos \: 20^{\circ} - \dfrac{1}{2} \: sin \: 20^{\circ}}{\dfrac{sin\:20^{\circ} \: cos \: 20^{\circ}}{2}}$

Replacing the values√3/2 and 1/2 with sin 60° and cos 60°

$\implies \dfrac{sin \: 60^{\circ} \: cos \: 20^{\circ} - cos \: 60^{\circ}\: sin \: 20^{\circ}}{\dfrac{sin\:20^{\circ} \: cos \: 20^{\circ}}{2}}$

Using Identity sin (A - B) = sin A cos B - cos A sin B

$\implies \dfrac{sin \: (60^{\circ} - 20^{\circ})} {\dfrac{1}{2} \:sin\:20^{\circ} \: cos \: 20^{\circ}}$

Multiplying this equation by 2/2

$\implies \dfrac{2}{2} \times 2 \times \dfrac{sin \: (60^{\circ} - 20^{\circ})}{sin\:20^{\circ} \: cos \: 20^{\circ}}$

$\implies 4 \times \dfrac{sin \: (60^{\circ} - 20^{\circ})}{2sin\:20^{\circ} \: cos \: 20^{\circ}}$

Using identity, sin 2A = 2 sin A cos A

$\implies 4 \times \dfrac{sin\: {40}^{\circ}}{sin\: {40}^{\circ}}$

4 = RHS

Hence proved.

Answer 2

Answer===  

Prove that  

Solving LHS,

Dividing numerator and denominator by 2

Replacing the values√3/2 and 1/2 with sin 60° and cos 60°

Using Identity sin (A - B) = sin A cos B - cos A sin B

Multiplying this equation by 2/2

Using identity, sin 2A = 2 sin A cos A

4 = RHS

Hence proved.