1.

if `(cosalpha)/(cosbeta)=a and (sinalpha)/(sinbeta)=b`, then the value of `sin^(2)beta` in terms of a and b isA. `(a^(2+1))/(a^(2)-b^(2))`B. `(a^(2)-b^(2))/(a^(2)+b^(2))`C. `(a^(2)-1)/(a^(2)-b^(2))`D. `(a^(2)-1)/(a^(2)+b^(2))`

Answer» Correct Answer - c
`(cos alpha)/(cos beta)=arArr cos alpha=a cos beta`
On squaring both sides
`cos^(2)alpha=a^(2)cos^(2)beta`
`rArr1-sin^(2)alpha=a^(2)(1-sin^(2)beta)` .........i
Again `sin alpha=6sin beta`
Squaring both sides
`rARrsin^(2)alpha=b^(2)sin^(2)beta`
put the value of `sin^(2)alpha` in equation (i)
`1-b^(2)sin^(2)beta=a^(2)-a^(2)sin^(2)beta`
`a^(2)-1=a^(2)sin^(2)beta-b^(2)sin^(2)beta`
`a^(2)-1=sin^(2)beta(a^(2)-b^(2))`
`rArr sin^(2)beta=(a^(2)-1)/(a^(2)-b^( 2))`


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