1.

If `0^(@)anglethetaangle90^(@)`, then find the value of `theta` from the equation `(cos^(2)theta)/(cot^(2)theta-cos^(2)theta)=3`.

Answer» `(cos^(2)theta)/(cot^(2)theta-cos^(2)theta)=3`
`rArr cos^(2)theta/((cos^(2)theta)/(sin^(2)theta)-cos^(2)theta)=3rArr(cos^(2)theta)/(cos^(2)theta((1)/(sin^(2)theta)-))=3`
` rArr(1)/(1/(sin^(2)theta)-1)=3rArr(1/(1-sin^(2)theta)=3)/(sin^(2)theta)`
`rArr(sin^(2)theta)/(1-sin^(2)theta)=3rArr(sin^(2)theta)/(cos^(2)theta)=3`
`rArrtan^(2)theta=(sqrt(3))^(2)`
`rArrtantheta=sqrt(3)` (taking positive sign only)
`rArrtantheta=tan60^(@)`
`rArrtheta=60^(@)`


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