Question

solve this question and prove also​

Answer 1

$\huge \mathfrak{solution}$

Given:

$x = p \: sec \theta + q \:tan \theta \: - - - - - (1)$

and

$y = p \tan \theta + q \sec \theta - - - - - (2)$

To prove:

$\bold{x ^{2} - {y}^{2} = {p}^{2} - {q}^{2} }$

Square and subtract both the equation 1) and 2)

$\rightarrow \: {x}^{2} - {y}^{2} = \\ ( {p \sec \theta + q \tan \theta)}^{2} - ( {p \tan \theta + q \sec \theta)}^{2} \\ \\ \rightarrow \: {x}^{2} - {y}^{2} = \\ {p}^{2} {sec}^{2} \theta + {q}^{2} {tan }^{2} \theta - 2p \: sec \theta \: q \: tan \theta \\ - ( {p}^{2} {tan}^{2} \theta + {q}^{2} {sec}^{2} \theta + 2p \: tan \theta \: q sec \theta$

Here ,

• 2 pq sec@tan@ gets cancel out

$\rightarrow \: {x}^{2} - {y}^{2} = \\ {p}^{2} {sec}^{2} \theta + {q}^{2} {tan}^{2} \theta - {p}^{2} {tan }^{2} \theta - {q}^{2} {sec}^{2} \theta \\ \\ \rightarrow \: x {}^{2} - {y}^{2} = \\ {p}^{2} ( {sec}^{2} \theta - {tan}^{2} \theta) - {q}^{2} ( {sec}^{2} \theta - {tan}^{2} \theta$

As we know that :

• sec^2 @ - tan^2@ =1

$\rightarrow \: {x}^{2} - {y}^{2} = {p}^{2} - {q}^{2}$

$\huge \boxed{proved}$