Question

If sec x + tan x = p, obtain the values of sec x, tan x and sin x in terms of p

Answer 1

Answer:

$\bold\red{\sec(x) = \frac{ {p}^{2} + 1}{2p}} \\ \\ \bold\red{\tan(x) = \frac{ {p}^{2} - 1}{2p} }\\ \\ \bold\red{\sin(x) = \frac{ {p}^{2} - 1}{ {p}^{2} + 1 } }$

Step-by-step explanation:

Given,

$\sec(x) + \tan(x) = p$

Therefore,

$\sec(x) - \tan(x) = \frac{1}{p}$

Therefore,

we get,

$2 \sec(x) = p + \frac{1}{p} \\ \\ = > \sec(x) = \frac{ {p}^{2} + 1 }{2p}$

Thus,

we get,

$\tan(x) = p - \frac{ {p}^{2} + 1}{2p} = \frac{2 {p}^{2} - {p}^{2} - 1}{2p} \\ \\ = > \tan(x) = \frac{ {p}^{2} - 1 }{2p}$

But,

we know that,

$\frac{ \tan(x) }{ \sec(x) } = \sin(x) \\ \\ = \sin(x) = \frac{ \frac{ {p}^{2} - 1}{2p} }{ \frac{ {p}^{2} + 1 }{2p} } \\ \\ = > \sin(x) = \frac{ {p}^{2} - 1 }{ {p}^{2} + 1}$

Hence,

$\bold{\sec(x) = \frac{ {p}^{2} + 1}{2p}} \\ \\ \bold{\tan(x) = \frac{ {p}^{2} - 1}{2p} }\\ \\ \bold{\sin(x) = \frac{ {p}^{2} - 1}{ {p}^{2} + 1 } }$