Question

please give me answer of maths question 6​

Answer 1

$\huge\bold{QUESTION}$

♡If Secθ+Tanθ=m, Then Prove that

Secθ=$= \dfrac{ {m}^{2} + 1 }{2m}$

$\huge\bold{ANSWER}$

Given:-

❥Secθ+Tanθ=m________(i)

To Prove:-

❥$Secθ = \dfrac{ {m}^{2} + 1 }{2m}$

Proof:-

According To An Identity,

❥Sec²θ-Tan²θ=1________(ii)

We can Factorise Eqⁿ (ii)

[a²-b²=(a+b)(a-b)]

$= > ( secθ + tanθ)(secθ - tanθ) = 1 \\ \\ = > m(secθ - tanθ) = 1$

(Since, Secθ+Tanθ=m)

$= > Secθ - Tanθ= \dfrac{1}{m}$________(iii)

❥From Eqⁿ (i) &(iii)

$\: \: \: \: Secθ+Tanθ=m \\ + Secθ - Tanθ= \dfrac{1}{m}$

_________________________

$= > 2Secθ + 0=m + \dfrac{1}{m} \\ \\ = > Secθ = \frac{ {m}^{2} +1 }{2m}$

. ˙ . ❥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¯¹ x

$\; \: \: \; \;$╬HOPE IT HELPS╬

Answer 2

Answer:

\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¯¹