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प्रश्न
`(sec^2 theta -1)(cosec^2 theta - 1)=1`
उत्तर
LHS = `(sec^2 theta -1)(cosec^2 theta-1)`
=`tan^2 theta xx cot^2 theta ( ∵ sec^2 theta - tan^2 theta = 1 and cosec^2 theta - cot^2 theta =1)`
=` tan^2 theta xx1/(cos^2theta)`
=1
=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`
