Question

Verify that:
${\cot }^2{60}^o+{\sin }^2{45}^o+{\sin }^2{30}^o+{\cos }^2{90}^o=\frac{13}{12}$

Answer 1

L.H.S$={\left(\frac{1}{\sqrt{3}}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{2}\right)}^2+0$

$=\frac{1}{3}+\frac{1}{2}+\frac{1}{4}+0$

L.C.M$(2,3,4)=12$

$=\frac{1}{3}\times \frac{4}{4}+\frac{1}{2}\times \frac{6}{6}+\frac{1}{4}\times \frac{3}{3}$

$=\frac{4+6+3}{12}$

$=\frac{13}{12}=$R.H.S