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प्रश्न
If `sin θ = 3/4` prove that `sqrt(("cosec"^2θ - cot^2θ)/(sec^2θ - 1)) = sqrt(7)/3`.
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उत्तर
We have `sin theta = 3/4`
In ΔABC
`AC^2 = AB^2 + BC^2`
`=> (4)^2 = (3)^2 + BC^2`
`=> BC^2= 16 - 9`
`=> BC^2 = 7`
`=> BC = sqrt7`
`:. cosec theta = 4/3, sec theta = 4/sqrt7 and cot theta = sqrt7/3`
Now
L.H.S `sqrt((cosec^2 theta - cot^2 theta)/(sec^2 theta - 1))`
`= sqrt(((4/3)^2 - (sqrt7/3)^2)/((4/sqrt7)^2 - 1)`
`= sqrt((16/9 - 7/9)/(16/7 - 1)`
`=sqrt((9/9)/((16 - 7)/7 ))`
`= sqrt(7/9)`
`= sqrt7/3`
= R.H.S
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