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प्रश्न
Prove that (cosec A – sin A)(sec A – cos A) sec2 A = tan A.
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उत्तर
L.H.S = `(cosec A - sin A)(secA - cosA)sec^2A`
`= (1/sinA - sinA)(1/cosA - cosA)(1/cos^2A)`
`= ((1 - sin^2A)/sin A)((1- cos^2A)/cos A)(1/(cos^2A))`
`= cos^2A/sinA . sin^2A/cos A . 1/cos^2A`
`= sinA/cosA`
= tan A
= R.H.S
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"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`
∴ `sin^2 theta + cos^2 theta = square`
