Question #6987d

3 Answers
Cem Sentin
Dec 2, 2017

Explanation below

Explanation:

#m=csc(theta)-sin(theta)#

=#1/sin(theta)-sin(theta)#

=#[1-(sin(theta))^2]/sin(theta)#

=#(cos(theta))^2/sin(theta)#

#n=sec(theta)-cos(theta)#

=#=1/cos(theta)-cos(theta)#

=#[1-(cos(theta))^2]/cos(theta)#

=#(sin(theta))^2/cos(theta)#

Hence,

#(m^2*n)^(2/3)+(n^2*m)^(2/3)#

=#[(cos(theta))^4/(sin(theta))^2*(sin(theta))^2/cos(theta)]^(2/3)+[(sin(theta))^4/(cos(theta))^2*(cos(theta))^2/sin(theta)]^(2/3)#

=#((cos(theta))^3)^(2/3)+((sin(theta))^3)^(2/3)#

=#cos(theta)^((3*2/3)#+#sin(theta)^((3*2/3)#

=#(cos(theta))^2+(sin(theta))^2#

=#1#

P dilip_k
Dec 2, 2017

Given

#csctheta-sintheta=m#

#=>1/sintheta-sintheta=m#

#=>(1-sin^2theta)/sintheta=m#

#=>cos^2theta/sintheta=m#

Again

#sectheta-cos theta=n#

#=>1/costheta-cos theta=n#

#=>(1-cos^2theta)/cos theta=n#

#=>sin^2theta/cos theta=n#

Now
#(m^2n)^(2/3 )+ (mn^2)^(2/3)#

#=((cos^2theta/sintheta)^2sin^2theta/costheta)^(2/3 )+ (cos^2theta/sintheta(sin^2theta/costheta)^2)^(2/3)#

#=(cos^3theta)^(2/3 )+ (sin^3theta)^(2/3)#

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

Ratnaker Mehta
Feb 11, 2018

Kindly refer to a Proof in the Explanation.

Explanation:

Here is a Third Method to Prove the Assertion :

#m=csctheta-sintheta=1/sintheta-sintheta#,

#:. m=(1-sin^2theta)/sintheta=cos^2theta/sintheta............(1)#.

Similarly, #n=sin^2theta/costheta............................(2)#.

#:. mn=sinthetacostheta..............................(3)#.

Also, #m/n=cos^3theta/sin^3theta:. (m/n)^(1/3)=costheta/sintheta.....(4)#.

#:. (3)xx(4) rArr mn*(m/n)^(1/3)=sinthetacostheta*costheta/sintheta, i.e., #

#m^(4/3)n^(2/3)=cos^2theta#.

Similarly, #m^(2/3)n^(4/3)=sin^2theta#, and, adding these,

#m^(4/3)n^(2/3)+m^(2/3)n^(4/3)=1,# or, what is the same as,

# (m^2n)^(2/3)+(mn^2)^(2/3)=1#.

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