Question

How do you prove `(sinx + cosx)(tanx + cotx)=secx + cscx`?

Answer 1

`color(green)[(sinx + cosx)(tanx + cotx)=secx-cosx+cosx+sinx+cscx-sinx=secx+cscx]`

Explanation:

show below:

`color(blue)[(sinx + cosx)(tanx + cotx)=secx + cscx]`

`L.H.S=color(blue)[(sinx + cosx)(tanx + cotx)]=`

`sinx*tanx+sinx*cotx+cosx*tanx+cosx*cotx=`

`sin^2x/cosx+cosx+sinx+cos^2x/sinx=`

`(1-cos^2x)/cosx+cosx+sinx+(1-sin^2x)/sinx=`

`1/cosx-cos^2x/cosx+cosx+sinx+1/sinx-sin^2x/sinx=`

`secx-cosx+cosx+sinx+cscx-sinx=color(blue)[secx+cscx]=R.H.S`

`color(red)["Useful Trigonometric Identities"]`

`cos^2theta+sin^2theta=1`

`1+tan^2theta=sec^2theta`

`sin2theta=2sin theta cos theta`

`cos2theta=cos^2theta-sin^2theta=2cos^2theta-1=1-2sin^2theta`

`cos^2theta=1/2(1+cos2theta)`

`sin^2theta=1/2(1-cos2theta)`

`tanx=sinx/cosx`

`cotx=cosx/sinx`

`1/cosx=secx`

`1/sinx=cscx`

Answer 2

Please see below.

Explanation:

We know that,

`color(red)((1)tantheta=sintheta/costheta and cottheta=costheta/sintheta`

`color(blue)((2)sin^2theta+cos^2theta=1`

`color(violet)((3)1/sintheta=csctheta and 1/costheta=sectheta`

We have to prove,

`(sinx+cosx)(tanx+cotx)=secx+cscx`

We take Left Hand Side :

`LHS=(sinx+cosx)(tanx+cotx)...tocolor(red)(Apply(1)`

`LHS=(sinx+cosx)(sinx/cosx+cosx/sinx)`

`LHS=(sinx+cosx)((sin^2x+cos^2x)/(sinxcosx))`

`LHS=(sinx+cosx)(1/(sinxcosx))...tocolor(blue)(Apply(2)`

`LHS=sinx/(sinxcosx)+cosx/(sinxcosx)`

`LHS=1/cosx+1/sinx...tocolor(violet)(Apply(3)`

`LHS=secx+cscx`

`LHS=RHS`