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प्रश्न
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
पर्याय
\[\frac{x^2 + 1}{x}\]
\[\frac{x^2 - 1}{x}\]
\[\frac{x^2 + 1}{2x}\]
\[\frac{x^2 - 1}{2x}\]
उत्तर
Given:
`sec θ+tanθ=x`
We know that,
`sec^2 θ-tan^2 θ=1`
⇒` (sec θ+tan θ)(sec θ-tanθ)=1`
⇒`x(sec θ-tan θ)=1`
⇒ `secθ-tan θ=1/x`
Now,
`secθ+tan θ=x,`
`sec θ-tan θ=1/x`
Subtracting the second equation from the first equation, we get
`(secθ+tan θ)-(secθ-tanθ)=x-1/x`
⇒` secθ+tanθ-secθ+tanθ=(x^2-1)/x`
⇒ `2 tanθ=(x^2-1)/x`
⇒ `2 tan θ=(x^2-1)/(2x)`
⇒ `tan θ=(x^2-1)/(2x)`
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