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Question
For ΔABC , prove that :
`tan ((B + C)/2) = cot "A/2`
Solution
`tan ((B + C)/2) = cot "A/2`
We know that for a triangle ΔABC
`<A + <B + <C = 180^circ`
`<B + <C = 180^circ - <A`
`(<B + <C)/2 = 90^circ - (<A)/2`
`tan ((B + C)/2) = tan(90^circ - A/2)`
= `cot(A/2)`
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cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
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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
