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If secα=2/√3 , then find the value of (1−cosecα)/(1+cosecα) where α is in IV quadrant. - Geometry

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If `sec alpha=2/sqrt3`  , then find the value of `(1-cosecalpha)/(1+cosecalpha)` where α is in IV quadrant.

 
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Solution

 

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`

 
Concept: Trigonometric Identities
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