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Solution
Given , sin 21°= x/y . So , cos21°=$$\sqrt{1-sin^{2}21°}$$ . cos21=$$\sqrt{1-(\frac{x}{y})^{2}}$$
So, sec21°= $$\frac{1}{cos21^{\circ}}$$
Also sin69°=cos21°.
Hence, sec21°-sin69° = $$\frac{1}{\sqrt{1-(\frac{x}{y})^{2}}}$$ - $$\sqrt{1-(\frac{x}{y})^{2}}$$
= $$\frac{x^2}{y\sqrt {y^2-x^2}}$$
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