Question

Question #02bba

Answer 1

Given `0< u< pi/2` and `tanu=5/3`

we get `u=tan^-1(5/3)~~59^@`

So `2u=118^@->"In 2nd quadrant"`

Hence `sin2u->"+ve"`

But `tan2u and cos 2u ->"-ve"`

Now

`sin2u=(2tanu)/(1+tan^2u)`

`=>sin2u=(2xx5/3)/(1+(5/3)^2)=10/3xx9/34=15/17`

`cos2u=-(1-tan^2u)/(1+tan^2u)`

`=>cos2u=(1-(5/3)^2)/(1+(5/3)^2`

`=>cos2u=-16/9xx9/34-8/17`

`tan2u=(2tanu)/(1-tan^2u)`

`=>tan2u=(2xx5/3)/(1-(5/3)^2)=-10/3xx9/16=-15/8`