Question

Eliminate θ from the equations a sec θ − c tan θ = b and bsec θ + d tan θ = c.

Answer 1

Answer:

$(c^2+bd)^2-(b^2-ac)^2=(ad+bc)^2$

Step-by-step explanation:

Given equations are

$a\:sec\theta-c\:tan\theta-b=0$

$b\:sec\theta+d\:tan\theta-c=0$

By cross multiplication rule

$\frac{sec\theta}{c^2+bd}=\frac{tan\theta}{-b^2+ac}=\frac{1}{ad+bc}$

$\implies\:\frac{sec\theta}{c^2+bd}=\frac{1}{ad+bc}$

$\implies\:sec\theta=\frac{c^2+bd}{ad+bc}$

and

$\frac{tan\theta}{-b^2+ac}=\frac{1}{ad+bc}$

$tan\theta=\frac{-b^2+ac}{ad+bc}$

we know that

$\boxed{sec^2\theta-tan^2\theta=1}$

$\implies\:(\frac{c^2+bd}{ad+bc})^2-(\frac{-b^2+ac}{ad+bc})^2=1$

$\implies\:\frac{(c^2+bd)^2}{(ad+bc)^2}-\frac{(-b^2+ac)^2}{(ad+bc)^2}=1$

$\implies\:(c^2+bd)^2-(-b^2+ac)^2=(ad+bc)^2$

$\implies\:\boxed{(c^2+bd)^2-(b^2-ac)^2=(ad+bc)^2}$