Leave a Comment / trigonometry / By

 

Q)  If ~x = a~ sec~\theta + b ~tan~\theta ~and~y = a~ tan~\theta + b~ sec~\theta, ~prove~that~ x ^ 2 - y ^ 2 = a ^ 2 - b ^ 2

Ans:  Here, x^2-y^2 = (a~ sec~\theta + b ~tan~\theta)^2 - (a~ tan~\theta + b~ sec~\theta)^2

= (a^2\sec{^2\theta}+2ab~ sec~\theta ~tan~\theta+b^2\tan{^2\theta}) – (a^2\tan{^2\theta}+2ab~ sec~\theta ~tan~\theta+b^2\sec{^2\theta}

= a^2\sec{^2\theta}+2ab~ sec~\theta ~tan~\theta+b^2\tan{^2\theta} – a^2\tan{^2\theta}-2ab~ sec~\theta ~tan~\theta-b^2\sec{^2\theta}

= a^2\sec{^2\theta}+\cancel{2ab~ sec~\theta ~tan~\theta}+b^2\tan{^2\theta} – a^2\tan{^2\theta}-\cancel{2ab~ sec~\theta ~tan~\theta}-b^2\sec{^2\theta}

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

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

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

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

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

We~know~that~ (\sec{^2\theta}-\tan{^2\theta})=1

\therefore \bf{x^2-y^2=a^2-b^2}...Hence~ Proved !

Please do press “Heart” button if you liked the solution. 

Related Posts

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top