Question

If `u=sec2theta+tan2theta`, find `(u-1)/(u+1)`?

Answer 1

Please see below.

Explanation:

We are given `u=sec2theta+tan2theta`....(1)

and hence using componendo and dividendo

`(u-1)/(u+1)=(sec2theta+tan2theta-1)/(sec2theta+tan2theta+1)`

multiplying numerator and denominator on RHS by `cos2theta`, we get

`(u-1)/(u+1)=(1+sin2theta-cos2theta)/(1+sin2theta+cos2theta)`

= `(1+2sinthetacostheta-cos^2theta+sin^2theta)/(1+2sinthetacostheta+cos^2theta-sin^2theta)`

`=(sin^2theta+2sinthetacostheta+sin^2theta)/(cos^2theta+2sinthetacostheta+cos^2theta)`

`=(2sin^2theta+2sinthetacostheta)/(2cos^2theta+2sinthetacostheta)`

`=(2sintheta(sintheta+costheta))/(2costheta(costheta+sintheta)`

= `(2sintheta)/(2costheta)`

= `tantheta`

= `t`

Answer 2

`tantheta`.

Explanation:

Prerequisite : Componendo Dividendo (CD) :

`CD : a/b=c/d iff (a-b)/(a+b)=(c-d)/(c+d)`.

We have, `u=sec2theta+tan2theta`,

`=1/(cos2theta)+(sin2theta)/(cos2theta)`,

`=(1+sin2theta)/(cos2theta)`,

`={(cos^2theta+sin^2theta)+2costhetasintheta}/(cos^2theta-sin^2theta)`,

`=(costheta+sintheta)^cancel2/{cancel((costheta+sintheta))(costheta-sintheta)}`,

`rArr u/1=(costheta+sintheta)/(costheta-sintheta)`.

`"By CD, then, "(u-1)/(u+1)`,

`={(costheta+sintheta)-(costheta-sintheta)}/{(costheta+sintheta)+(costheta-sintheta)}`,

`=(2sintheta)/(2costheta)`,

`rArr (u-1)/(u+1)=tantheta`,

as Respected Shwetank Mauria has readily derived!

Answer 3

`tantheta.` Kindly go through Explanation.

Explanation:

Here is a Third Method to solve the Problem, if we use,

`cos2theta=(1-t^2)/(1+t^2), and, tan2theta=(2t)/(1-t^2), t=tantheta`.

Hence, `u=sec2theta+tan2theta=1/(cos2theta)+tan2theta`,

`=(1+t^2)/(1-t^2)+(2t)/(1-t^2)`,

`=(1+2t+t^2)/(1-t^2)`,

`=(1+t)^2/{(1+t)(1-t)}`.

`rArr u/1=(1+t)/(1-t)`.

`rArr (u-1)/(u+1)=(2t)/(2)=t=tantheta`, as Before!