If ${tan}^{2} θ = 1 - k^{2}$ $tan^2theta=1-k^2$ , then prove that $sec θ + {tan}^{3} θ csc θ = {(2 - k^{2})}^{\frac{3}{2}}$ $sectheta+tan^3thetacsctheta=(2-k^2)^(3/2)$ ?

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Answer 1 · 2017-02-11T18:17:27

Please see below.

As ${tan}^{2} θ = 1 - k^{2}$ $tan^2theta=1-k^2$

$sec θ + {tan}^{3} θ csc θ$ $sectheta+tan^3thetacsctheta$

= $sec θ + {tan}^{2} θ . \frac{sin θ}{cos θ} \frac{.1}{sin θ}$ $sectheta+tan^2theta.sintheta/costheta.1/sintheta$

= $sec θ + {tan}^{2} θ . sec θ$ $sectheta+tan^2theta.sectheta$

= $sec θ (1 + {tan}^{2} θ)$ $sectheta(1+tan^2theta)$

= ${sec}^{3} θ$ $sec^3theta$

= ${({sec}^{2} θ)}^{\frac{3}{2}}$ $(sec^2theta)^(3/2)$

= ${(1 + {tan}^{2} θ)}^{\frac{3}{2}}$ $(1+tan^2theta)^(3/2)$

= ${(1 + 1 - k^{2})}^{\frac{3}{2}}$ $(1+1-k^2)^(3/2)$

= ${(2 - k^{2})}^{\frac{3}{2}}$ $(2-k^2)^(3/2)$

If tan2θ=1−k2tan^2theta=1-k^2, then prove that secθ+tan3θcscθ=(2−k2)32sectheta+tan^3thetacsctheta=(2-k^2)^(3/2)?