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Question
Prove that `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1
Solution
`"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")`
= `((cos "A")/(sin "A"))/(1 - (sin "A")/(cos "A")) + ((sin "A")/(cos "A"))/(1 - (cos "A")/(sin "A"))`
= `((cos "A")/(sin "A"))/((cos "A" - sin "A")/(cos "A")) + ((sin "A")/(cos "A"))/((sin "A" - cos "A")/(sin "A"))`
= `"cos A"/"sin A" xx "cos A"/(cos "A" - sin "A") + "sin A"/"cos A" xx "sin A"/(sin "A" - cos "A")`
= `(cos^2"A")/(sin "A"(cos "A" - sin "A")) + (sin^2"A")/(cos"A"(sin"A" - cos"A"))`
= `1/(sin "A" - cos "A") ((-cos^3"A" + sin^3"A")/(sin"A" cos"A"))`
= `1/(sin"A" - cos"A")((sin^3"A" - cos^3"A")/(sin"A" cos"A"))`
= `1/(sin"A" - cos"A")xx ((sin"A" - cos"A")(sin^2"A" + sin"A" cos"A" + cos^2"A"))/(sin"A" cos"A")` ......[∵ a3 – b3 = (a – b)(a2 + ab + b2)]
= `(sin^2"A" +sin"A" cos"A" + cos^2"A")/(sin"A" cos"A"` ......(i)
= `(1 + sin"A" cos"A")/(sin"A" cos"A")` .....[∵ sin2A + cos2A = 1]
= `1/(sin"A" cos"A") + (sin"A" cos"A")/(sin"A" cos"A")`
= cosec A sec A + 1 .....(ii)
`"cot A"/(1 - tan "A") + "tan A"/(1 - cot "A")`
= `(sin^2"A" + sin"A" cos"A" + cos^2"A")/(sin"A" cos"A")` ......[From (i)]
= `(sin^2"A")/(sin"A" cos"A") + "sin A cos A"/"sin A cos A" + (cos^2"A")/"sin A cos A"`
= `"sin A"/"cos A" + 1 + "cos A"/"sin A"`
= tan A + 1 + cot A ......(iii)
From (ii) and (iii), we get
`"cot A"/(1 - tan "A") + "tan A"/(1 - cot "A")` = 1 + tan A + cot A = sec A . cosec A + 1
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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
