Question

Prove that $b^2{x}^2-a^2{y}^2=a^2{b}^2,$ if:
i) $x=a\sec \theta ,y=btan\theta$
ii) $x=a\cos ec\theta ,y=b\cot \theta$

Answer 1

$b^2{x}^2-a^2{y}^2=a^2{b}^2$

i) $x=a\sec \theta ,y=b\tan \theta$

$LHS=b^2{x}^2-a^2{y}^2$

$=b^2{a}^2{\sec }^2\theta -a^2{b}^2{\tan }^2\theta$

$=a^2{b}^2({\sec }^2\theta -{\tan }^2\theta )$

$=a^2{b}^2=RHS$

$\because {\sin }^2+{\cos }^2\theta =1$

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

Dividing both sides by ${\cos }^2\theta$

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

$(\frac{\sin \theta }{\cos \theta }=\tan \theta )$

ii) $a\csc \theta =x$

$b\cot \theta =y$

$LHS=b^2{x}^2-a^2{y}^2$

$=b^2{a}^2{\csc }^2\theta -a^2{b}^2{\cot }^2\theta$

$=a^2{b}^2({\csc }^2\theta -{\cot }^2\theta )$

$=a^2{b}^2$

$=RHS$

$\because {\csc }^2\theta$

$-{\cot }^2\theta$

$=1$