If \( \sin 21^\circ = \frac{x}{y} \), then find \( \sec 21^\circ – \sin 69^\circ \) is equal to.

Solution

Given:
\[ \sin 21^\circ = \frac{x}{y} \]

We need to find the value of \(\sec 21^\circ – \sin 69^\circ\).

Using the identity \(\sin(90^\circ – \theta) = \cos \theta\), we have:
\[ \sin 69^\circ = \sin(90^\circ – 21^\circ) = \cos 21^\circ \]

Now, we know that \(\cos 21^\circ = \sqrt{1 – \sin^2 21^\circ} = \sqrt{1 – \left(\frac{x}{y}\right)^2} = \frac{\sqrt{y^2 – x^2}}{y}\).

Therefore, the expression \(\sec 21^\circ – \sin 69^\circ\) becomes:
\[ \sec 21^\circ – \sin 69^\circ = \frac{1}{\cos 21^\circ} – \cos 21^\circ \]
\[ = \frac{y}{\sqrt{y^2 – x^2}} – \frac{\sqrt{y^2 – x^2}}{y} \]
\[ = \frac{y^2 – (y^2 – x^2)}{y\sqrt{y^2 – x^2}} \]
\[ = \frac{x^2}{y\sqrt{y^2 – x^2}} \]

Conclusion

The value of \(\sec 21^\circ – \sin 69^\circ\) is \(\frac{x^2}{y\sqrt{y^2 – x^2}}\).