Question

Prove:
[tex]\sqrt{3} cosec 20° - sec 20° = 4
Wrong answers will be reported.

Answer 1

Answer:

Formula used :-

$\rm \green{\frac{1}{sin \theta} = cosec \theta }$

$\rm\blue{ \frac{1}{cos \theta} = sec \theta}$

$\rm \orange{sinA.cosB - cosA.sinB = sin(A - B)}$

$\rm \purple{sinA.cosB - cosA.sinB = sin(A - B)}$

Step-by-step explanation:

$\rm \sqrt{3} .cosec 20° - sec 20° = 4$

$\rm LHS= \sqrt{3} .cosec 20° - sec 20° = 4$

$\scriptsize \rm = \frac{ \sqrt{3} }{sin20°} - \frac{1}{cos20°} \: [ \because \: \frac{1}{sin \theta} = cosec \theta \: and \: \frac{1}{cos \theta} = sec \theta]$

$\rm = \frac{ \sqrt{3}.cos20° - \: sin20°}{sin20°cos20°}$

$\rm On \: multiplying \: the \: numerator \: by \: 2 \times \frac{1}{2} ,$

$\rm = \frac{ 2 \times \frac{1}{2} [\sqrt{3}.cos20° - \: sin20° ]}{sin20°cos20°}$

$\rm = \frac{ 2 [ \frac{ \sqrt{3} }{2} .cos20° - \: \frac{1}{2}. sin20° ]}{sin20°cos20°}$

$\rm = \frac{ 2 [ sin60°.cos20° - \: cos60°. sin20° ]}{sin20°cos20°}$

$\scriptsize \rm = \frac{2sin(60° - \: 20°)}{sin20°.cos20°} \: [ \because \: sinA.cosB - cosA.sinB = sin(A - B)]$

$\rm = \frac{2sin40°}{sin20°.cos20°}$

On multiplying the numerator by 2 in the denominator,

$\rm = \frac{2(2sin40°)}{2(sin20°.cos20°)}$

$\rm = \frac{4sin40°}{2sin20°.cos20°}$

$\rm = \frac{4sin40°}{sin2(20°)} \: [ \because \: 2sinA.cosA = sin2A]$

$\rm = \frac{4 \cancel{sin40°}}{ \cancel{sin40°}}$

$= 4$

$\rm = RHS$