1.

यदि `tanA=ntanB` और `sinA=m,sinB` तो `cos^(2)A` का मान है?A. `(m^(2)+1)/(n^(2)+1)`B. `(m^(2)+1)/(n^(2)-1)`C. `(m^(2)-1)/(n^(2)-1)`D. `(m^(2)-1)/(n^(2)+1)`

Answer» Correct Answer - c
`sinA =msinB`
`sin^(2)A=m^(2)sin^(2)B`........(i)
Now, `tan^(2)A=n^(2)tan^(2)B`
`(sin^(2)A)/(cos^(2)A)=n^(2)(sin^(2)B)/(cos^(2)B)`
from equation (i)
`(1-cos^(2)A)/(n^(2)cos^(2)A)=((1-cos^(2)A))/((m^(2))/(1-(sin^(2)A)/(m^(2))))`
`(1-cos^(2)A)/(n^(2)cos^(2)A)=(1-cos^(2)A)/(m^(2)-1+cos^(2)A) `
`rArrm^(2)-1+cos^(2)A=n^(2)cos^(2)A`
`m^(2)-1=cos^(2)theta(n^(2)-1)`
`cos^(2)A=(m^(2)-1)/(n^(2)-1)`


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