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प्रश्न
`sqrt((1 + sin θ)/(1 - sin θ)) = sec θ + tan θ`
Prove the following:
`sqrt((1 + sin θ)/(1 - sin θ)) = (sec θ + tan θ)`
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उत्तर
LHS = `sqrt((1 + sin θ)/(1 - sin θ))`
=`sqrt(((1 + sin θ))/(1 - sin θ) xx ((1 + sin θ))/(1 + sin θ))`
=` sqrt(((1 + sin θ)^2)/(1 - sin^2 θ))`
=`sqrt(((1 + sin θ)^2)/(cos^2 θ))`
=`(1 + sin θ)/cos θ`
=`1/cos θ + (sin θ)/(cos θ)`
= sec θ + tan θ
= RHS
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
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∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
