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Question
If sec θ + tan θ = x, then sec θ =
Options
\[\frac{x^2 + 1}{x}\]
\[\frac{x^2 + 1}{2x}\]
\[\frac{x^2 - 1}{2x}\]
\[\frac{x^2 - 1}{x}\]
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Solution
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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Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is ______.
