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Question
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
Solution
LHS = mn
`= (cosec theta + cot theta) (cosec theta - cot theta)`
`= cosece^2 theta - cot^2 theta`
= 1 [∵ `(1 + b)(a - b) = a^2 - b^2 cosec^2 theta - cot^2 theta = 1`]
=RHS
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Activity:
L.H.S = `square`
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= `(cos^2theta + sin^2theta)/square`
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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
