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प्रश्न
Prove that:
(tan A + cot A) (cosec A – sin A) (sec A – cos A) = 1
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उत्तर
(tan A + cot A) (cosec A – sin A) (sec A – cos A)
= `(sinA/cosA + cosA/sinA)(1/sinA - sinA)(1/cosA - cosA)`
= `((sin^2A + cos^2A)/(sinAcosA))((1 - sin^2A)/sinA)((1 - cos^2A)/cosA)`
= `(1/(sinAcosA))(cos^2A/sinA)(sin^2A/cosA)`
= 1
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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`
