Question

the value of sin theta tan theta/1-cos theta+tan^2 theta- sec^2theta is​

Answer 1

Given:$\[\frac{{\sin \theta \tan \theta }}{{1 - \cos \theta }} + {\tan ^2}\theta - {\sec ^2}\theta \]$.

To find: the value of the given expression.

Solution:

Simplify the given expression.

$\[\frac{{\sin \theta \tan \theta }}{{1 - \cos \theta }} + {\tan ^2}\theta - {\sec ^2}\theta \]$

$\[\left[ {\tan \theta = \frac{{\sin \theta }}{{\cos \theta }}} \right]\]$

$\[\frac{{\sin \theta \tan \theta }}{{1 - \cos \theta }} + {\tan ^2}\theta - {\sec ^2}\theta = \frac{{\sin \theta \frac{{\sin \theta }}{{\cos \theta }}}}{{1 - \cos \theta }} + \left( { - \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)} \right)\]$

Apply, $\[\left[ {\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right) = 1} \right]\]$

                                     $\[ = \frac{{\sin \theta \frac{{\sin \theta }}{{\cos \theta }}}}{{1 - \cos \theta }} + \left( { - 1} \right)\]$

                                     $\[ = \frac{{{{\sin }^2}\theta }}{{\cos \theta \left( {1 - \cos \theta } \right)}} + \left( { - 1} \right)\]$

                                    $\[ = \frac{{{{\sin }^2}\theta }}{{\cos \theta - {{\cos }^2}\theta }} + \left( { - 1} \right)\]$

Take LCM of the above expression.

                                   $\[ = \frac{{{{\sin }^2}\theta - 1\left( {\cos \theta - {{\cos }^2}\theta } \right)}}{{\cos \theta - {{\cos }^2}\theta }}\]$

                                   $\[ = \frac{{{{\sin }^2}\theta - \cos \theta + {{\cos }^2}\theta }}{{\cos \theta - {{\cos }^2}\theta }}\]$

                                   $\[ = \frac{{{{\sin }^2}\theta + {{\cos }^2}\theta - \cos \theta }}{{\cos \theta - {{\cos }^2}\theta }}\]$

Apply, $\[\left[ {{{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right]\]$

                                   $\[ = \frac{{1 - \cos \theta }}{{\cos \theta - {{\cos }^2}\theta }}\]$

                                   $\[ = \frac{{1 - \cos \theta }}{{\cos \theta \left( {1 - \cos \theta } \right)}}\]$

                                   $\[ = \frac{1}{{\cos \theta }}\]$

                                   $\[ = \sec \theta \]$

hence, the value of $\[\frac{{\sin \theta \tan \theta }}{{1 - \cos \theta }} + {\tan ^2}\theta - {\sec ^2}\theta \]$ is $\[\sec \theta \]$.

Answer 2

thanks

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