Question

the slant height of a frustum of a com

Answer 1

$\sqrt{ \frac{ \sec( \alpha ) - 1}{ \sec( \alpha ) + 1} } + \sqrt{ \frac{ \sec( \alpha ) + 1 }{ \sec( \alpha ) - 1} } \\ = \frac{ \sqrt{ {( \sec( \alpha ) - 1) }^{2} } + \sqrt{ {( \sec( \alpha ) + 1) }^{2} } }{ \sqrt{( \sec( \alpha) - 1)( \sec( \alpha ) + 1) } } \\ = \frac{( \sec( \alpha ) - 1) + ( \sec( \alpha ) + 1)}{ \sqrt{ { \sec( \alpha ) }^{2} - 1 } } \\ = \frac{2 \sec( \alpha ) }{ \sqrt{ { \tan( \alpha ) }^{2} } } \\ = \frac{2 \sec( \alpha ) }{ \tan( \alpha ) } \\ = \frac{2 \frac{1}{ \cos( \alpha ) } }{ \frac{ \sin( \alpha ) }{ \cos( \alpha ) } } \\ = \frac{2}{ \sin( \alpha ) } \\ = 2 \csc( \alpha ) proved$

Answer 2

The aalttachment will help you ( hope so)

And sorry for the mistake in middle