If `tantheta=t t h e nt a n2theta+sec2theta` is equal to `(1+t)/(1-t)` (b) `(1-t)/(1+t)` (c) `(2t)/(1-t)` (d) `(2t)/(1+t)`

Explanation

Given,

`tantheta=t->(i)`

we know that `tan2theta=(2tantheta)/(1-tan^2theta)`

and `sec2theta=(1+tan^2theta)/(1+tan^2theta)`

now,

`=tan2theta+sec2theta`

`=(2tantheta)/(1-tan^2theta)+(1+tan^2theta)/(1+tan^2theta)`

`=(2t)/(1-t^2)+(1+t^2)/(1-t^2)`[using(i)]

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