Question

Plz answer in step by step procedure....
Points will be awarded.... Thanks!

Answer 1

Answer:

$\frac{p\:sin\theta-q\:cos\theta}{p\:sin\theta-q\:cos\theta}=\frac{p^2-q^2}{p^2+q^2}$

Step-by-step explanation:

Given:

$sec\theta=\frac{\sqrt{p^2+q^2}}{q}$

$cos\theta=\frac{q}{\sqrt{p^2+q^2}}$

We know that

$\boxed{sin^2\theta=1-cos^2\theta}$

$sin^2\theta=1-\frac{q^2}{p^2+q^2}$

$sin^2\theta=\frac{p^2+q^2-q^2}{p^2+q^2}$

$sin^2\theta=\frac{p^2}{p^2+q^2}$

$\implies\:sin\theta=\frac{p}{\sqrt{p^2+q^2}}$

Now,

$\frac{p\:sin\theta-q\:cos\theta}{p\:sin\theta-q\:cos\theta}$

$=\frac{p(\frac{p}{\sqrt{p^2+q^2}})-q(\frac{q}{\sqrt{p^2+q^2}})}{p(\frac{p}{\sqrt{p^2+q^2}})-q(\frac{q}{\sqrt{p^2+q^2}})}$

$=\frac{\frac{p^2}{\sqrt{p^2+q^2}}-\frac{q^2}{\sqrt{p^2+q^2}}}{\frac{p^2}{\sqrt{p^2+q^2}}+\frac{q^2}{\sqrt{p^2+q^2}}}$

$=\frac{\frac{p^2-q^2}{\sqrt{p^2+q^2}}}{\frac{p^2+q^2}{\sqrt{p^2+q^2}}}$

$=\frac{p^2-q^2}{p^2+q^2}$

$\implies\:\boxed{\frac{p\:sin\theta-q\:cos\theta}{p\:sin\theta-q\:cos\theta}=\frac{p^2-q^2}{p^2+q^2}}$