Without using trigonometric tables, find the value of the following expression:
$\frac{sec( 90^{o}-\theta ) .cosec\theta -tan( 90^{o} -\theta ) cot\theta +cos^{2} 25^{o}+cos^{2} 65^{o} }{3tan27^{o} tan63^{o}}$.

$\frac{sec( 90^{o} -\theta ) .cosec\theta -tan( 90^{o} -\theta ) cot\theta +cos^{2} 25^{o} +cos^{2} 65^{o} }{3tan27^{o} tan63^{o}}$

To do: To find the value of the given expression without using trigonometric tables.

Solution: The given experession :

$\frac{sec( 90^{o} -\theta ) .cosec\theta -tan( 90^{o} -\theta ) cot\theta +cos^{2} 25^{o} +cos^{2} 65^{o} }{3tan27^{o} tan63^{o}}$

 $\because sec( 90^{o} -\theta ) =cosec\theta ,\ sin( 90^{o} -\theta ) =cos\theta ,\ tan( 90^{o} -\theta ) =cot\theta $

$=\frac{cosec\theta .cosec\theta -cot\theta .cot\theta +sin^{2}( 90^{o}-25^{o} ) +cos^{2} 65^{o} }{3cot( 90^{o} -27^{o} ) .tan63^{o} }$

$=\frac{cosec^{2} \theta -cot^{2} \theta +sin^{2} 65^{o}+cos^{2} 65^{o} }{3cot63^{o} tan63^{o} }$ 

$=\frac{1+1}{3\times 1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \because cosec^{2} \theta -cot^{2} \theta =1,\ sin^{2} \theta +cos^{2} \theta =1\ and\ cot\theta .tan\theta =1\right)$ 

$=\frac{2}{3}$

The value of the given expression is $\frac{2}{3}$.