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प्रश्न
(i)` (1-cos^2 theta )cosec^2theta = 1`
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उत्तर
LHS= `(1-cos^2 theta) cosec^2 theta`
=`sin ^2 theta cosec^2 theta (∵ cos^2 theta + sin^2 theta =1)`
=`1/(cosec^2theta) ×cosec^2theta`
=1
Hence, LHS = RHS
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Activity:
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= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
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Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
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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
