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प्रश्न
If sec θ + tan θ = x, then sec θ =
पर्याय
\[\frac{x^2 + 1}{x}\]
\[\frac{x^2 + 1}{2x}\]
\[\frac{x^2 - 1}{2x}\]
\[\frac{x^2 - 1}{x}\]
उत्तर
Given: `sec θ+tan θ=1`
We know that,
`sec^2θ-tan^2θ=1`
⇒ `(secθ+tan θ)(secθ-tan θ)=1`
⇒`x(sec θ-tan θ)=1`
⇒ `secθ-tan θ=1/x`
Now,
`sec θ+tan =x`
`sec θ-tan θ=1/x`
Adding the two equations, we get
`(sec θ+tan θ)+(sec θ-tan θ)=x+1/x`
⇒` sec θ+tan θ+sec θ-tan θ=(x^2+1)/x`
⇒ `2 sec θ=(x^2+1)/x`
⇒` sec θ=(x^2+1)/(2x)`
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