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प्रश्न
Prove that
`cot^2A-cot^2B=(cos^2A-cos^2B)/(sin^2Asin^2B)=cosec^2A-cosec^2B`
उत्तर
`cot^2A-cot^2B`
`=cos^2A/sin^2A-cos^2B/sin^2B`
`=(cos^2Asin^2B-cos^2Bsin^2A)/(sin^2Asin^2B)`
`=(cos^2A(1-cos^2B)-cos^2B(1-cos^2A))/(sin^2Asin^2B)`
`=(cos^2A-cos^2Acos^2B-cos^2B+cos^2Bcos^2A)/(sin^2Asin^2B)`
`=(cos^2A-cos^2B)/(sin^2Asin^2B)`
`=(1-sin^2A-1+sin^2B)/(sin^2Asin^2B)`
`=(-sin^2A+sin^2B)/(sin^2Asin^2B)`
`=sin^2B/(sin^2AsinB)-sin^2A/(sin^2Asin^2B)`
`=1/sin^2A-1/sin^2B`
= cosec2A - cosec2B
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tan (55° + x) = cot (35° – x)
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
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`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
Choose the correct alternative:
cos 45° = ?
Prove that `(cos^2theta)/(sintheta) + sintheta` = cosec θ
Prove that `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
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`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
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