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प्रश्न
Prove the following identity :
`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`
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उत्तर
LHS = `(tanθ + secθ - 1)/(tanθ - secθ + 1) `
= `(tanθ + secθ - {sec^2θ - tan^2θ})/(1 + tanθ -secθ)`
= `(tanθ + secθ - {(secθ + tanθ)(secθ - tanθ)})/(1 + tanθ - secθ)`
= `([tanθ + secθ]{1 - (secθ - tanθ)})/[[1 + tanθ - secθ]` = `([tanθ + secθ][1 + tanθ - secθ])/[[1 + tanθ - secθ]]`
= `[tanθ + secθ] = (1 + sinθ)/cosθ` = RHS
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Show that, cotθ + tanθ = cosecθ × secθ
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
