Three quarter of a thousand

$\csc(75^\circ)=\frac{1}{sin(75^\circ)}$ -(1)

$\sin(75^\circ)=sin(45^\circ)*cos(30^\circ)+cos(45^\circ)*sin(30^\circ)$

we get $\sin(75^\circ)=\frac{\sqrt{6}+\sqrt{2}}{4}$

from(1)

$\csc(75^\circ)=\frac{4}{\sqrt{6}+\sqrt{2}}$

rationalizing $\frac{4}{\sqrt{6}+\sqrt{2}}=\sqrt{6}-\sqrt{2}$

hence proved:

$\csc(75^\circ)=\sqrt{6}-\sqrt{2}$

cosec(2x360+30)

cosec30=2

As we know.. Csc750° = csc30°

ans csc is the inverse of sin..

csc750°= csc30°= 1/sin30°

csc750°= csc30°= 2 ..

Csc=1/sin so we find sin (750) Sin750 =0.5 so csc (750) = 1/0.5 =2

750 degrees is the same as 720 degrees (2 full laps) and 30 degrees. We know from unit circle that the sin of 30 degrees is 1/2. For csc, just flip the fraction. Answer is 2.

Always keeping in the mind the phasor rotates with a period of 360 degrees. For ex. A minute hand in a clock ( assume it makes an angle ᶿ (theta) from position 12 in a clock ; when minute hand completes one rotation angle of 360 degrees. Similarly for two rotation it is 360 * 2=720 deg. ) So in this problem for any number greater than 360, reduce it to the form X=Number- 360*n ; where, n is an integer (1,2,3,......) in the above problem, x= 720 - 360 * 2; » x=30.

so finally csc(x)=1/sinx , sin(30)=1/2;

Ans. » csc(30)= 2

Cosec(750°)=cosec(720°+30°)=cosec30°=1/sin30°=2

750-2*360=30 Cosec30=1/sin30=2 + with ASTC