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प्रश्न
Prove the following identity :
`sec^2A.cosec^2A = tan^2A + cot^2A + 2`
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उत्तर
LHS = `sec^2A.cosec^2A = 1/(cos^2A.sin^2A)`
RHS = `tan^2A + cot^2A + 2 = tan^2A + cot^2A + 2tan^2A.cot^2A`
= `(tanA + cotA)^2 = (sinA/cosA + cosA/sinA)^2`
= `((sin^2A + cos^2A)/(sinA.cosA))^2 = 1/(cos^2A.sin^2A)`
= Hence , LHS = RHS
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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`
