secx-tanx=x find cosx,sinx,tanx
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Answer:
cos x =2x /x^2+1
sinx=x^2-1
Given :
To find :
Solution :
Let ,
[tex] \tt \: \sec(x) - \tan(x) = x \longrightarrow \: eq(1)[/tex]
using the identity ,
[tex] {\boxed{ \bigstar {\rm{ \: | \sec {}^{2} (x) - { \tan }^{2} (x) = 1}}}}[/tex]
[tex] : \implies \tt \: \sec {}^{2} (x) - \tan {}^{2} (x) = 1[/tex]
Now ,
[tex] {\boxed{\rm{ \bigstar{ \: a {}^{2} - {b}^{2} = (a + b)(a - b)}}}}[/tex]
[tex] : \implies \tt [\sec(x) - \tan(x) ][ \sec(x) + \tan(x) ]= 1[/tex]
From eq(1) ,
[tex] : \implies \tt \: (x)[\sec(x) + \tan(x) ] = 1 \\ \\ : \implies \tt \: [ \sec(x) + \tan(x)] = \frac{1}{x} \longrightarrow \: eq(2)[/tex]
Adding eq(1) & eq(2) ,
[tex] \implies \tt \: \sec(x) - \tan(x) = x \\ \\ \: \: \: \: \tt \: \: \: \: \: \: \: \: \: \sec(x) + \tan(x) = \frac{1}{x} \\ \: \: \: \: \: \: \: \: \: \: - - - - - - - - - - \\ \tt \: \: \: \: \: \: \: \: \: \: \: 2 \sec(x) = \frac{1}{x} + x \\ \\ \: \: \: \: \: \: \: \: \: \implies \tt 2 \sec(x) = \frac{1 + {x}^{2} }{x} \\ \\ \: \: \: \: \: \: \: \: \: \implies \tt \sec(x) = \frac{1 + x {}^{2} }{2x} [/tex]
From eq(2) ,
[tex] : \implies \tt \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg) + \tan(x) = \dfrac{1}{x} \\ \\ \tt : \implies \tan(x) = \frac{1}{x} - \bigg( \frac{1 + {x}^{2} }{2x} \bigg) \\ \\ : \implies \tt \tan(x) = \frac{2 - 1 - {x}^{2} }{2x} \\ \\ \tt : \implies \: \tan(x) = \frac{1 - x {}^{2} }{2x} \\ \\ : \implies \tt \tan(x) = \frac{1 - {x}^{2} }{2x} [/tex]
Now ,
[tex] : \implies \tt \: \dfrac{1 }{ \cos(x) } = \sec(x) = \dfrac{1 + x {}^{2} }{2x} \\ \\ \tt : \implies \: \frac{1}{ \cos(x) } = \frac{1 + {x}^{2} }{2x} \\ \\ \tt: \implies \: 2x = [1 + {x}^{2} ] \cos(x) \\ \\ : \tt \implies \cos(x) = \frac{2x}{(1 + {x}^{2}) } [/tex]
The value of Sin(x) is ,
[tex] \tt : \implies \: \dfrac{ \sin(x) }{ \cos(x) } = \tan(x) = \frac{1 - {x}^{2} }{2x} \\ \\ \tt : \implies \: \frac{ \sin(x) }{ \cos(x) } = \frac{1 - {x}^{2} }{2x} \\ \\ : \implies \tt \: \frac{ \sin(x) }{ ( \frac{2x}{1 + {x}^{2} }) } = \frac{1 - {x}^{2} }{2x} \\ \\ : \implies \tt \: 2x[\sin(x) ] = \bigg(\frac{2x}{1 + {x}^{2} } \bigg)(1 - {x}^{2} ) \\ \\ : \tt\implies \: \sin(x) = \frac{1 - {x}^{2} }{1 + {x}^{2} } [/tex]
∴ [tex] \sf \dfrac{2x}{1 + {x}^{2} } , \dfrac{1 - {x}^{2} }{2x} \: and \: \dfrac{1 - {x}^{2} }{1 + x {}^{2} } [/tex]
are the values of Cos(x) , Tan(x) & Sin(x)