Question

If sec x + tan x = p. Find value of sec x and tan x in terms of p.​

Answer 1

$\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{secx + tanx = p} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{secx}\\ &\sf{tanx}\end{cases}\end{gathered}\end{gathered}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:secx + tanx = p - - - (1)$

We know that,

$\rm :\longmapsto\: {sec}^{2} x - {tan}^{2} x = 1$

$\rm :\longmapsto\:(secx + tanx)(secx - tanx) = 1$

$\rm :\longmapsto\:p(secx - tanx) = 1 \: \: \: \: \: \: \: \{ \: using \: (1) \}$

$\rm :\implies\:secx - tanx = \dfrac{1}{p} - - - (2)$

On adding equation (1) and equation (2), we get

$\rm :\longmapsto\:2secx = p + \dfrac{1}{p}$

$\rm :\longmapsto\:2secx = \dfrac{ {p}^{2} + 1}{p}$

$\rm :\implies\: \boxed{\bf\:secx = \dfrac{ {p}^{2} + 1}{2p} }$

On Subtracting equation (2) from equation (1), we get

$\rm :\longmapsto\:2tanx = p - \dfrac{1}{p}$

$\rm :\longmapsto\:2tanx = \dfrac{ {p}^{2} - 1}{p}$

$\rm :\implies\: \boxed{\bf\:tanx = \dfrac{ {p}^{2} - 1}{2p}}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1