Question

The value of ${\sec }^2{45}^{\circ }-{\tan }^2{45}^{\circ }$ is equal to

• A. ${\sin }^2{60}^{\circ }-{\cos }^2{60}^{\circ }$
• B. ${\sin }^2{45}^{\circ }+{\cos }^2{60}^{\circ }$
• C. ${\sec }^2{60}^{\circ }-{\tan }^2{60}^{\circ }$
• D. $0$

Answer 1

The correct option is C ${\sec }^2{60}^{\circ }-{\tan }^2{60}^{\circ }$

${\sec }^{2}45°-{\tan }^{2}45°={\left(\sqrt{2}\right)}^2-{\left(1\right)}^2=2-1=1$

Now, $\left(A\right){\sin }^{2}60°-{\cos }^{2}60°={\left(\frac{\sqrt{3}}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$

$\left(B\right){\sin }^{2}45°+{\cos }^{2}60°={\left(\frac{1}{\sqrt{2}}\right)}^2-{\left(\frac{1}{2}\right)}^2=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$

$\left(C\right){\sec }^{2}60°-{\tan }^{2}60°={\left(2\right)}^2-{\left(\sqrt{3}\right)}^2=4-3=1$

$\therefore {\sec }^{2}45°-{\tan }^{2}45°={\sec }^{2}60°-{\tan }^{2}60°$

Hence, the answer is ${\sec }^{2}60°-{\tan }^{2}60°.$