Question

Prove that 1÷sin10"-√3÷cos10"

Answer 1

1/ sin 10 - √3/cos 10
= [1*cos 10 - √3 sin10]/sin10cos10

multiply numerator and denomenator with 1/2
= [ 1/2cos 10 - √3/2sin 10]/[1/2 * sin10cos10]
=[sin30cos10 -cos30sin10]/[2/4* sin10cos10]

= sin(30-10)/[1/4 *sin(2*10)] {∵sinAcosB- cosAsinB = sin(A-B)]
=4*sin20/sin20 { ∵2sinX cosX = sin2X}
= 4