Question

if x=p tan +q. sec ∅ and y=psec∅+q tan ∅ then prove that x^2-y^2=q^2-p^2​

Answer 1

Question:

$\sf if~x=p ~tan \theta +q~sec \theta ~and ~y=p~sec \theta +q~tan \theta, ~then~ prove ~that~ x^2-y^2=q^2-p^2$

Solution:

According to the question:

We have:

• $\sf x=p ~tan \theta +q~sec \theta$
• $\sf y=p~sec \theta +q~tan \theta$

Now,

$\sf x^2-y^2 = \Big(p ~tan \theta +q~sec \theta \Big)^2 - \Big(p ~sec \theta +q~tan \theta \Big)^2 \\\\ \sf Using~\boxed{(a+b)^2=a^2+b^2+2ab}, we~have: \\\\ \sf \Big(p^2tan^2 \theta +q^2sec^2 \theta +2pq~tan \theta sec \theta \Big)- \Big(p^2sec^2 \theta +q^2tan^2 \theta +2pq~sec \theta tan \theta \Big) \\\\ \sf \Big(p^2tan^2 \theta +q^2sec^2 \theta +2pq~tan \theta sec \theta - p^2sec^2 \theta - q^2tan^2 \theta - 2pq~sec \theta tan \theta \Big) \\\\ \sf \Big(p^2tan^2 \theta +q^2sec^2 \theta + \cancel{2pq~tan \theta sec \theta} - p^2sec^2 \theta - q^2tan^2 \theta - \cancel{2pq~sec \theta tan \theta} \Big) \\\\ \sf \Big(p^2tan^2 \theta +q^2sec^2 \theta - p^2sec^2 \theta - q^2tan^2 \theta \Big) \\\\ \sf \Big(p^2tan^2 \theta - p^2sec^2 \theta +q^2sec^2 \theta - q^2tan^2 \theta \Big)$

Taking $\sf p^2$ and $\sf q^2$, common:

$\sf \Big[p^2(tan^2 \theta - sec^2 \theta) +q^2(sec^2 \theta - tan^2 \theta) \Big] \\\\ \sf \implies \Big[p^2(-1) +q^2(1) \Big] \\\\ \implies \sf - p^2+q^2 \\\\ \implies \sf q^2-p^2$

Hence, proved.

Extras:

$\begin{gathered}\sf Trigonometric \: Identities \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ } 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}$