Question

Sin 30 + tan 45 - cosec 60 is divided by sec 30 + cos 60 + cot 45

Answer 1

$\frac{\sin30+\tan45-\csc60}{\sec30+\cos60+\cot45}=\frac{\sin30+\tan45-\frac{1}{\sin60}}{\frac1{\cos30}+\cos60+\cot45}=\frac{0.5+1-\frac{2}{\sqrt3}}{\frac{2}{\sqrt3}+0.5+1}=\\\\\frac{(\frac32-\frac{2}{\sqrt3})^2}{(\frac32+\frac{2}{\sqrt3})(\frac32-\frac{2}{\sqrt3})}=\frac{(\frac32-\frac2{\sqrt3})^2}{(\frac32)^2-(\frac{2}{\sqrt3})^2}=\frac{\frac94-\frac{6}{\sqrt3}+\frac43}{\frac94-\frac43}=\frac{\frac{43}{12}-2\sqrt3}{\frac{11}{12}}=\\\\\frac{\frac{43}{12}-\frac{24\sqrt3}{12}}{\frac{11}{12}}=\frac{43-24\sqrt3}{11}\approx0.13$

Answer 2

hope it is correct.......