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 $tan^2theta=1-k^2$

$sectheta+tan^3thetacsctheta$

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

= $sectheta+tan^2theta.sectheta$

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

= $sec^3theta$

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

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

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

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

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