please give me answer of maths question 5

Answers
Step-by-step explanation:
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\huge\bold{QUESTION}QUESTION
♡If Secθ+Tanθ=m, Then Prove that
Secθ== \dfrac{ {m}^{2} + 1 }{2m}=
2m
m
2
+1
\huge\bold{ANSWER}ANSWER
Given:-
❥Secθ+Tanθ=m________(i)
To Prove:-
❥Secθ = \dfrac{ {m}^{2} + 1 }{2m}Secθ=
2m
m
2
+1
Proof:-
According To An Identity,
❥Sec²θ-Tan²θ=1________(ii)
We can Factorise Eqⁿ (ii)
[a²-b²=(a+b)(a-b)]
\begin{gathered} = > ( secθ + tanθ)(secθ - tanθ) = 1 \\ \\ = > m(secθ - tanθ) = 1\end{gathered}
=>(secθ+tanθ)(secθ−tanθ)=1
=>m(secθ−tanθ)=1
(Since, Secθ+Tanθ=m)
= > Secθ - Tanθ= \dfrac{1}{m}=>Secθ−Tanθ=
m
1
________(iii)
❥From Eqⁿ (i) &(iii)
\begin{gathered} \: \: \: \: Secθ+Tanθ=m \\ + Secθ - Tanθ= \dfrac{1}{m} \end{gathered}
Secθ+Tanθ=m
+Secθ−Tanθ=
m
1
_________________________
\begin{gathered} = > 2Secθ + 0=m + \dfrac{1}{m} \\ \\ = > Secθ = \frac{ {m}^{2} +1 }{2m} \end{gathered}
=>2Secθ+0=m+
m
1
=>Secθ=
2m
m
2
+1
. ˙ . ❥PROVED
ADDITIONAL INFORMATION
❥Some Basic Formulas
Sin²x+Cos²x=1
Sec²x-Tan²x=1
Cosec²x-Cot²x=1
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
sin(x+y) = sin(x)cos(y)+cos(x)sin(y)
cos(x+y) = cos(x)cos(y)–sin(x)sin(y)
sin(x–y) = sin(x)cos(y)–cos(x)sin(y)
cos(x–y) = cos(x)cos(y) + sin(x)sin(y)
sin¯¹ (–x) = – sin¯¹ x
cos¯¹ (–x) = π – cos¯¹x
tan¯¹ (–x) = – tan¯¹ x
cosec¯¹(–x) = – cosec¯¹ x
sec¯¹ (–x) = π – sec¯¹ x
cot¯¹(–x) = π – cot¯