Question

Without using trigonometric tables,prove that $\left(7to11\right)$:

• A. ${\sin }^2{28}^{\circ }-{\cos }^2{62}^{\circ }=0$
• B. ${\cos }^2{25}^{\circ }+{\cos }^2{65}^{\circ }=1$
• C. $\cos e{c}^2{67}^{\circ }-{\tan }^2{23}^{\circ }=1$
• D. $se{c}^2{22}^{\circ }-{\cot }^2{68}^{\circ }=1$

Answer 1

The correct option is C $\cos e{c}^2{67}^{\circ }-{\tan }^2{23}^{\circ }=1$

$\left(a\right){\sin }^2{28}^{\circ }-{\cos }^2{62}^{\circ }$

$={\sin }^2{28}^{\circ }-{\cos }^2({90}^{\circ }-{28}^{\circ })$

$={\sin }^2{28}^{\circ }-{\sin }^2{28}^{\circ }=0$ is proved.

$\left(b\right){\cos }^2{25}^{\circ }+{\cos }^2{65}^{\circ }$

$={\cos }^2{25}^{\circ }+{\cos }^2({90}^{\circ }-{25}^{\circ })$

$={\cos }^2{25}^{\circ }+{\sin }^2{25}^{\circ }=1$ is proved.

$\left(c\right){\csc }^2{67}^{\circ }-{\tan }^2{23}^{\circ }$

$={\csc }^2{67}^{\circ }-{\tan }^2({90}^{\circ }-{67}^{\circ })$

$={\csc }^2{67}^{\circ }-{\cot }^2{67}^{\circ }=1$ is proved.

$\left(d\right){\sec }^2{22}^{\circ }-{\cot }^2{68}^{\circ }$

$={\sec }^2{22}^{\circ }-{\cot }^2({90}^{\circ }-{22}^{\circ })$

$={\sec }^2{22}^{\circ }-{\tan }^2{22}^{\circ }=1$ is proved.