Question

prove that 1/sin 10° - √3/ cos 10°= 4

Answer 1

hey!! here is your answer for the Question...



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


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
hope it helps you... plz mark as brainliest...

Answer 2

Hello!

$\dfrac{1}{\sf sin(10)} - \dfrac{\sqrt{3}}{\sf cos(10)} \\ \\ \implies 4 \times \dfrac{\dfrac{1}{2} \sf cos(10)- \dfrac{\sqrt{3}}{2} \sf sin(10)}{\sf 2\: sin(10).cos(10)} \\ \\ \implies 4 \times \dfrac{\sf sin(30).cos(10) - cos(30).sin(10)}{\sf 2\:sin(10).cos(10)} \\ \\ \implies 4 \times \dfrac{\cancel{\sin(20)}}{\cancel{\sin(20)}} \implies \boxed{\red{\bf{4}}}$

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