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प्रश्न
If sec θ = x + `1/(4"x"), x ≠ 0,` find (sec θ + tan θ)
उत्तर
Given: Sec θ = x + `1/(4"x"), x ≠ 0,`
Squaring both sides
sec2 θ = `("x" + 1/(4"x"))^2`
We know
tan2θ = sec2θ - 1
`=>"tan"^2theta = ("x" + 1/("4x"))^2 - 1`
`=> "tan"^2theta = ("x" + 1/"4x" - 1)("x" + 1/"4x" + 1)`
`=> "tan"^2theta = ("x" - 1/("4x"))^2`
`=> "tan" theta = +- ("x" -1/("4x"))`
When tan θ = x - `1/("4x")`
secθ + tan θ = x + `1/("4x") + "x" - 1/("4x")`
= 2x
When tan θ = - `("x" - 1/("4x")) = 1/("4x") - "x"`
sec θ + tan θ = x + `1/"4x" + 1/"4x" - "x"`
`=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
