Question

Without using trigonometric tables, evaluate ${\text{cosec}}^2{57}^0-{\tan }^2{33}^0+\cos {44}^0\text{cosec}{46}^0-\sqrt{2}\cos {45}^0-{\tan }^2{60}^0$

Answer 1

${\text{cosec}}^2{57}^0-{\tan }^2{33}^0+\cos {44}^0\text{cosec}{46}^0-\sqrt{2}\cos {45}^0-{\tan }^2{60}^0$

$={\text{cosec}}^2{57}^0-{\tan }^2(90-{57}^0)+\cos (90-46{)}^0\text{cosec}{46}^0-\sqrt{2}(\frac{1}{\sqrt{2}})-(\sqrt{3}{)}^2$

$={\text{cosec}}^2{57}^0-co{t}^2{57}^0+\sin {46}^0\text{cosec}{46}^0-1-3$

$=1+\sin {46}^0\left(\frac{1}{\sin {46}^0}\right)-1-3$

$=1+1-1-3=-2$