#### Question

if `sin theta = 3/4` prove that `sqrt(cosec^2 theta - cot)/(sec^2 theta - 1) = sqrt7/3`

#### Solution

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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#### APPEARS IN

Solution If Sin Theta = 3/4 Prove that `Sqrt(Cosec^2 Theta - Cot)/(Sec^2 Theta - 1) = Sqrt7/3` Concept: Trigonometric Ratios.