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Question
Prove that cot2θ × sec2θ = cot2θ + 1.
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Solution
L.H.S. = cot2θ × sec2θ
= `(cos^2θ)/(sin^2θ) xx 1/(cos^2θ)`
= `1/(sin^2θ)`
= cosec2θ
= 1 + cot2θ ...[∵ 1 + cot2θ = cosec2θ]
= R.H.S.
∴ cot2θ × sec2θ = cot2θ + 1
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To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
