[tex]\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{secx + tanx = p} \end{cases}\end{gathered}\end{gathered}[/tex] [tex]\begin{gathered}\begin{gathered}\bf \: To \: Find – \begin{cases} &\sf{secx}\\ &\sf{tanx}\end{cases}\end{gathered}\end{gathered}[/tex] [tex]\large\underline{\sf{Solution-}}[/tex] Given that, [tex]\rm :\longmapsto\:secx + tanx = p – – – (1)[/tex] We know that, [tex]\rm :\longmapsto\: {sec}^{2} x – {tan}^{2} x = 1[/tex] [tex]\rm :\longmapsto\:(secx + tanx)(secx – tanx) = 1[/tex] [tex]\rm :\longmapsto\:p(secx – tanx) = 1 \: \: \: \: \: \: \: \{ \: using \: (1) \}[/tex] [tex]\rm :\implies\:secx – tanx = \dfrac{1}{p} – – – (2) [/tex] On adding equation (1) and equation (2), we get [tex]\rm :\longmapsto\:2secx = p + \dfrac{1}{p} [/tex] [tex]\rm :\longmapsto\:2secx = \dfrac{ {p}^{2} + 1}{p} [/tex] [tex]\rm :\implies\: \boxed{\bf\:secx = \dfrac{ {p}^{2} + 1}{2p} }[/tex] On Subtracting equation (2) from equation (1), we get [tex]\rm :\longmapsto\:2tanx = p – \dfrac{1}{p} [/tex] [tex]\rm :\longmapsto\:2tanx = \dfrac{ {p}^{2} – 1}{p} [/tex] [tex]\rm :\implies\: \boxed{\bf\:tanx = \dfrac{ {p}^{2} – 1}{2p}} [/tex] 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 Reply
[tex]\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{secx + tanx = p} \end{cases}\end{gathered}\end{gathered}[/tex]
[tex]\begin{gathered}\begin{gathered}\bf \: To \: Find – \begin{cases} &\sf{secx}\\ &\sf{tanx}\end{cases}\end{gathered}\end{gathered}[/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\rm :\longmapsto\:secx + tanx = p – – – (1)[/tex]
We know that,
[tex]\rm :\longmapsto\: {sec}^{2} x – {tan}^{2} x = 1[/tex]
[tex]\rm :\longmapsto\:(secx + tanx)(secx – tanx) = 1[/tex]
[tex]\rm :\longmapsto\:p(secx – tanx) = 1 \: \: \: \: \: \: \: \{ \: using \: (1) \}[/tex]
[tex]\rm :\implies\:secx – tanx = \dfrac{1}{p} – – – (2) [/tex]
On adding equation (1) and equation (2), we get
[tex]\rm :\longmapsto\:2secx = p + \dfrac{1}{p} [/tex]
[tex]\rm :\longmapsto\:2secx = \dfrac{ {p}^{2} + 1}{p} [/tex]
[tex]\rm :\implies\: \boxed{\bf\:secx = \dfrac{ {p}^{2} + 1}{2p} }[/tex]
On Subtracting equation (2) from equation (1), we get
[tex]\rm :\longmapsto\:2tanx = p – \dfrac{1}{p} [/tex]
[tex]\rm :\longmapsto\:2tanx = \dfrac{ {p}^{2} – 1}{p} [/tex]
[tex]\rm :\implies\: \boxed{\bf\:tanx = \dfrac{ {p}^{2} – 1}{2p}} [/tex]
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