Question

If $\sec \theta =\frac{\sqrt{p^2+q^2}}{q}$, then the value of $\frac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }$

• A. $\frac{p}{q}$
• B. $\frac{p^2}{q^2}$
• C. $\frac{p^2-q^2}{p^2+q^2}$
• D. $\frac{p^2+q^2}{p^2-q^2}$

Answer 1

The correct option is C $\frac{p^2-q^2}{p^2+q^2}$

$\sec \theta =\frac{hypotenuse}{base}=\frac{\sqrt{p^2+q^2}}{q}$
Hence,
$hypotenuse=\sqrt{p^2+q^2}$
$base=q$
Therefore by applying, Pythagoras theorem, we get the altitude as
$p$.
Hence
$\sin \theta =\frac{p}{\sqrt{p^2+q^2}}$
$\cos \theta =\frac{q}{\sqrt{p^2+q^2}}$
Hence substituting, the values in the above question we get
$\frac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }$
$=\frac{p^2-q^2}{p^2+q^2}$