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

3 Answers
Dec 13, 2016

Answer:

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#

Jun 19, 2018

Answer:

# 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!

Jun 19, 2018

Answer:

# 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!