Question

show that 1/sin10-√3/cos10=4

Answer 1

Formulas used :-
(1). sin(A) cos(B) - cos(A) sin(B) = sin(A - B)
(2). 2 sin(A) cos(A) = sin(2A)

Proof :-

$\frac{1}{sin(10)} - \frac{ \sqrt{3} }{cos(10)} \\ \\ 4 [ \frac{ \frac{1}{2} cos(10) - \frac{ \sqrt{3} }{2} sin(10) }{2 sin(10) cos(10)} ] \\ \\ 4[ \frac{sin(30) cos(10) - cos(30) sin(10)}{2 sin(10) cos(10)} ] \\ \\ 4[ \frac{sin(20)}{sin(20)} ] \\ \\ 4$

Answer 2

1/sin10-√3/cos10
= (cos10 - √3sin10)/cos10sin10
multiplying 1/2 in denominator and numerator
= {(cos10/2) - (√3sin10/2)}/(cos10sin10/2)
= (cos60cos10 - sin60sin10)/(cos10sin10/2)
= cos(60 + 10)/(cos10sin10/2)
= cos70/(cos10sin10/2)
= 2cos70/cos10sin10
multiplying 2 on numerator and denominator
= 4cos70/2sin10cos10
=4cos70/sin20
=4cos70/cos70
= 4  proved