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प्रश्न
Prove that cot2θ × sec2θ = cot2θ + 1
उत्तर
L.H.S = cot2θ × sec2θ
= `(cos^2theta)/(sin^2theta) xx 1/(cos^2theta)`
= `1/(sin^2theta)`
= cosec2θ
= 1 + cot2θ ......[∵ 1 + cot2θ = cosec2θ]
= R.H.S
∴ cot2θ × sec2θ = cot2θ + 1
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
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`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
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