Question : If $\sin 23^{\circ}=\frac{a}{b}$, the value of $\sec 23^{\circ}-\sin 67^{\circ}$ is:
Option 1: $\frac{a^2}{\sqrt{b^2-a^2}}$
Option 2: $\frac{b^2-a^2}{a b}$
Option 3: $\frac{a^2}{b\sqrt{b^2+a^2}}$
Option 4: $\frac{a^2}{b \sqrt{b^2-a^2}}$
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Correct Answer: $\frac{a^2}{b \sqrt{b^2-a^2}}$
Solution :
Given: $\sin 23^{\circ}=\frac{a}{b}$
To find: $\sec 23^{\circ}-\sin 67^{\circ}$
By using the figure in the form of a perpendicular, base, and hypotenuse.
$\sin 23^{\circ}=\frac{p}{h} =\frac{a}{b}$
So, base$=\sqrt{b^2-a^2}$
$\sec 23^{\circ} =\frac{b}{\sqrt{b^2-a^2}}$
$\sin 67^{\circ}=\frac{\sqrt{b^2-a^2}}{b}$
Putting the values, we get:
$\sec 23^{\circ}-\sin 67^{\circ}$ =$\frac{b}{\sqrt{b^2-a^2}}-\frac{\sqrt{b^2-a^2}}{b}$
= $\frac{a^2}{b \sqrt{b^2-a^2}}$
Hence, the correct answer is $\frac{a^2}{b \sqrt{b^2-a^2}}$.
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