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

2 Answers
May 28, 2018

#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#

May 28, 2018

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#