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

Feb 11, 2017

#### Explanation:

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

$\sec \theta + {\tan}^{3} \theta \csc \theta$

= $\sec \theta + {\tan}^{2} \theta . \sin \frac{\theta}{\cos} \theta \frac{.1}{\sin} \theta$

= $\sec \theta + {\tan}^{2} \theta . \sec \theta$

= $\sec \theta \left(1 + {\tan}^{2} \theta\right)$

= ${\sec}^{3} \theta$

= ${\left({\sec}^{2} \theta\right)}^{\frac{3}{2}}$

= ${\left(1 + {\tan}^{2} \theta\right)}^{\frac{3}{2}}$

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

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