If \(\sin 23^\circ=\displaystyle\frac{x}{y}\), then what is the value of \(\sec 23^\circ-\sin 67^\circ\)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{x^2}{y\sqrt{y^2-x^2}}\)
Since \(23^\circ+67^\circ=90^\circ\), \(\sin 67^\circ=\cos 23^\circ\). The expression then has only one angle, and \(\cos 23^\circ\) follows from \(\sin 23^\circ\); it is positive because \(23^\circ\) is an acute angle.
\(\sec 23^\circ-\sin 67^\circ\displaystyle=\frac{1}{\cos 23^\circ}-\cos 23^\circ=\frac{1-\cos^2 23^\circ}{\cos 23^\circ}=\frac{\sin^2 23^\circ}{\cos 23^\circ}\)
\(\cos 23^\circ\displaystyle=\sqrt{1-\frac{x^2}{y^2}}=\frac{\sqrt{y^2-x^2}}{y}\)
\(\displaystyle\frac{\sin^2 23^\circ}{\cos 23^\circ}=\frac{x^2}{y^2}\times\frac{y}{\sqrt{y^2-x^2}}=\frac{x^2}{y\sqrt{y^2-x^2}}\)