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प्रश्न
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
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उत्तर
L.H.S = sec2θ + cosec2θ
= 1 + tan2θ + 1 + cot2θ .....[∵ sec2θ = 1 + tan2θ and cosec2θ = 1 + cot2θ]
= 2 + tan2θ + cot2θ .....(i)
R.H.S = sec2θ x cosec2θ
= (1 + tan2θ) x (1 + cot2θ) .....[∵ sec2θ = 1 + tan2θ and cosec2θ = 1 + cot2θ]
= 1 + cot2θ + tan2θ + tan2θ x cot2θ
= 1 + cot2θ + tan2θ + tan2θ x (1/tan2θ) ...... [∵ cot2θ = 1/tan2θ]
= 2 + tan2θ + cot2θ .......(ii)
From (i) and (ii)
sec2θ + cosec2θ = sec2θ x cosec2θ
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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
