2*cos(59^circ*(31^circ*cosec)) %2B 3*cos(80^circ*(10^circ*cosec)) – InterviewSolution

1.	2cos(59^circ(31^circcosec)) %2B 3cos(80^circ(10^circcosec))
Answer» 3 cos80 cosec10+2 cos59 cosec31=3 cos80/ sin10+2 cos59/ sim31= 3 cos80_/sin(90-10) + 2 cos59/ sin(90-31)= 3 cos80/ cos80+2 cos59/ cos59=3+2=5

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