Question

${\sin }^{2}17.5°+{\sin }^{2}72.5°$ is equal to

• A. ${\cos }^{2}90°$
• B. ${\tan }^{2}45°$
• C. ${\cos }^{2}30°$
• D. ${\sin }^{2}45°$

Answer 1

The correct option is B

${\tan }^{2}45°$

Explanation for the correct answer:

Simplifying the given expression:

$$\begin{gathered} \Rightarrow {\sin }^{2}17.5°+{\sin }^{2}72.5° \\ \Rightarrow \sin 17.5°+{\sin }^2(90°–17.5°)\left[\because {\sin }^2(90-x)={\cos }^{2}x\right] \\ \Rightarrow {\sin }^{2}17.5°+{\cos }^{2}17.5° \\ \Rightarrow 1\left[\because \tan 45°=1\right] \\ \Rightarrow \tan 45° \\ \Rightarrow {\tan }^{2}45°\left[\because \tan 45°={\tan }^{2}45°=1\right] \end{gathered}$$

Therefore, the correct answer is option (B).