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प्रश्न
Prove that secθ + tanθ =`(costheta)/(1-sintheta)`.
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उत्तर
secθ + tanθ = `1/cosθ + sintheta/cosθ`
`=(1+sintheta)/costheta`
`=((1+sintheta)(1-sintheta))/(costheta (1-sintheta))`
`=(1^2 - sin^2theta)/(costheta(1-sintheta))`
`=cos^2theta/(costheta(1-sintheta))`
`therefore sectheta +tantheta =costheta/(1-sintheta)`
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In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
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