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प्रश्न
If `sec alpha=2/sqrt3` , then find the value of `(1-cosecalpha)/(1+cosecalpha)` where α is in IV quadrant.
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उत्तर
Given that α is in quadrant IV, where x is positive and y is negative.
`sec alpha=r/x=2/sqrt3`
`Let r=2k, `
`r^2=x^2+y^2`
`therefore(2k^2)=(sqrt(3k))^2+y^2`
`therefore y^2=4k^2-3k^2=k^2`
`therefore y=+-k`
`cosec alpha =r/y=(2k)/-k=-2`
Substituting the value of cosec ,we get
`(1-cosec alpha)/(1+cosec alpha)=(1-(-2))/(1+(-2))=(1+2)/(1-2)=3/-1 `
`(1-cosec alpha)/(1+cosec alpha)=-3`
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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`
