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प्रश्न
`(sin theta+1-cos theta)/(cos theta-1+sin theta) = (1+ sin theta)/(cos theta)`
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उत्तर
LHS= `(sin theta+1cos theta)/(cos theta-1+sin theta) `
=`((sin theta+1-cos theta)(sin theta+cos theta+1))/((cos theta -1 + sin theta)(sin theta + cos theta +1))`
=`((sin theta + 1 )^2 - cos^2 theta)/((sin theta + cos theta )^2 -1^2)`
=`(sin^2 theta +1+2 sin theta - cos^2 theta)/(sin^2 + cos^2 theta+2 sin theta cos theta -1)`
=`(sin^2 theta + sin^2 theta + cos^2 theta +2sin theta - cos^2 theta)/(2 sin theta cos theta)`
=`(2 sin ^2 theta + 2 sin theta)/(2 sin theta cos theta)`
=`(2 sin theta (1+ sin theta))/(2 sin theta cos theta)`
=`(1+sin theta)/cos theta`
= 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`
