If `sin A=a cos B and cosA=b sinB `then, `(a^2 - 1) tan^2 A+(1 -b^2)tan^2B` is equal to

Options

A. `(a^(2)-b^(2))/(b^(2))`
B. `(a^(2)-b^(2))/(a^(2))`
C. `(a^(2)+b^(2))/(b^(2))`
D. `(a^(2)+b^(2))/(a^(2))`

Answer

A

Explanation

We have,
` sinA=a cos B and cos A=bsinB`
`thereforea^(2)cos^(2)B+b^(2)sin^(2)B=sin^(2)A+cos^(2)A`
`implies a^(2)cos^(2)B+b^(2)(1-cos^(2)beta)=1`
`impliescos^(2)B=(1-b^(2))/(a^(2)-b^(2))and sin^(2)B=(a^(2)-1)/(a^(2)-b^(2))`
`impliestan^(2)B=(a^(2)-1)/(1-b^(2))`
Again, `sinA =a cos Band cos A=bsinB`
`impliestanA=a/bcotB`
`tan^(2)A=(a^(2))/(b^(2))((1-b^(2))/(a^(2)-1))`
`therefore(a^(2)-1)tan^(2)A+(1-b^(2))tan^(2)beta`
`=(a^(2))/(b^(2))(1-b^(2))+(a^(2)-1)=(a^(2)-b^(2))/(b^(2))`