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Question
`cot^2 theta - 1/(sin^2 theta ) = -1`a
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Solution
LHS = `cot^2 theta - 1/ (sin^2 theta)`
= `(cos^2 theta )/(sin^2 theta) - 1/(sin^2 theta)`
=`(cos^2 theta -1)/(sin^2 theta)`
=` (- sin^2 theta )/(sin ^2 theta)`
= -1
= 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
