Question

If+p=sec+tan+than+find+the+value+of+p^2-1/p^2+1

Answer 1

Answer:

To find :

$\implies \: \bf{\frac{ {p}^{2} - 1 }{ {p}^{2} + 1 } }$

Formula used :

$\star \: \boxed{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy} \\ \\ \star \: \boxed{ { \sec}^{2} \theta - { \tan }^{2} \theta = 1}$

Step-by-step explanation:

According to the question,

$\bf{p \: = \sec( \theta) + \tan( \theta) }$

Then ,

$\rightarrow \: \bf{\frac{ {p}^{2} - 1}{ {p}^{2} + 1 } } \\ \\ \bf{put \: the \: given \: value \: of \: p} \\ \\ \rightarrow \: \frac{ { \bigg(( \sec( \theta) + \tan( \theta) \bigg) }^{2} - 1 }{ { \bigg(\sec( \theta) + \tan( \theta) \bigg) }^{2} + 1} \\ \\ \red{\bf{using \: the \: given \: formula \: }} \\ \\ \rightarrow \: \small \: \frac{ { \sec}^{2} \theta + { \tan}^{2} \theta + 2 \sec( \theta) \tan( \theta) - 1 }{ \: \: \: { \sec}^{2} \theta + { \tan}^{2} \theta + 2 \sec( \theta) \tan( \theta) + 1 \: } \\ \\ \because { \sec }^{2} \theta - 1 = { \tan}^{2} \theta \: \: \\ \: \: \: \: \: \: 1 + { \tan }^{2} \theta = { \sec }^{2} \theta \\ \\ \rightarrow \: \frac{2 { \tan}^{2} \theta + \: 2 \sec( \theta) \tan( \theta)\: }{2 { \sec }^{2} \theta \: +2 \sec( \theta) \tan( \theta) \: }$

For next step please refer to the attachment,