1.

Prove that `1+tanA Tan(A/2)=tanA cot(A/2)-1=secA`.

Answer» Here, we will use, `tan2A = (2tanA)/(1-tan^2A)`
`1+tanAtan(A/2) = 1+((2tan(A/2))/(1-tan^2(A/2))*tan(A/2))`
`=(1-tan^2(A/2)+2tan^2(A/2))/(1-tan^2(A/2))`
`=(1+tan^2(A/2))/(1-tan^2(A/2))`
`= (sec^2(A/2))/((cos^2(A/2)-sin^2(A/2))/(cos^2(a/2)))`
`=(sec^2(A/2))/(sec^2(A/2)cosA)= secA`

Now, `tanAcot(A/2) -1 = ((2tan(A/2))/(1-tan^2(A/2)))(1/tan(a/2)) - 1`
`(2-1+tan^2(A/2))/(1-tan^2(A/2))`
`=(1+tan^2(A/2))/(1-tan^2(A/2))`
`= (sec^2(A/2))/((cos^2(A/2)-sin^2(A/2))/(cos^2(a/2)))`
`=(sec^2(A/2))/(sec^2(A/2)cosA)= secA`


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