If #tan^2theta=1-k^2#, then prove that #sectheta+tan^3thetacsctheta=(2-k^2)^(3/2)#?

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Answer 1 · 2017-02-11T18:17:27

Please see below.

As #tan^2theta=1-k^2#

#sectheta+tan^3thetacsctheta#

= #sectheta+tan^2theta.sintheta/costheta.1/sintheta#

= #sectheta+tan^2theta.sectheta#

= #sectheta(1+tan^2theta)#

= #sec^3theta#

= #(sec^2theta)^(3/2)#

= #(1+tan^2theta)^(3/2)#

= #(1+1-k^2)^(3/2)#

= #(2-k^2)^(3/2)#