Math, asked by charan8438, 10 months ago

If sec x +tan x=k, find the value of sin x

Answers

Answered by chandanpratik53
0

Answer:

sin x= \frac{k^{2}-1 }{k^{2}+1 } OR sin x= (k×cos x)-1

Step-by-step explanation:

We know that 1+tan²x=sec²x

⇒ ∴ sec²x-tan²x=1

⇒ (sec x+tan x)×(sec x-tan x)=1     [∵a²-b²=(a+b)(a-b)]

We know the value of (sec x+tan x) is k

⇒ ∴ (sec x-tan x)=\frac{1}{k}

Adding k and 1/k

k+\frac{1}{k}=(sec x+tan x)+(sec x-tan x)

\frac{k^{2}+1 }{k}=2 sec x

⇒ ∴ sec x=\frac{k^{2}+1 }{2k}

We know that cos x=\frac{1}{sec x}

⇒ ∴ cos x=\frac{2k}{k^{2}+1 }

⇒ (sec x+tan x) can be written as \frac{1}{cos x}+\frac{sin x}{cos x}

\frac{1}{cos x}+\frac{sin x}{cos x}=k

\frac{1+sin x}{cos x}=k

⇒ 1+sin x=k×cos x

Substituting value of (cos x),

⇒ 1+sin x=k×(\frac{2k}{k^{2}+1 })

⇒ sin x=\frac{2k^{2} }{k^{2}+1 } -1

⇒ sin x= \frac{2k^{2}-k^{2}-1  }{k^{2}+1 }

⇒ ∴ sin x= \frac{k^{2}-1 }{k^{2}+1 }

                                       OR

\frac{1}{cos x}+\frac{sin x}{cos x}=k

\frac{1+sin x}{cos x}=k

⇒ 1+sin x=k×cos x

⇒ ∴ sin x=(k×cos x)-1

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