Question

prove the identity (1+cot theta +tan theta )(sin theta - cos theta )=sec theta/cosec²theta-cosectheta/sec² theta

Answer 1

Here is the answer

Hope it helps

Answer 2

Given :

The Trigonometrical identity as

( 1 + cotФ + tanФ ) ( sinФ - cosФ ) = $\dfrac{sec\Theta }{cosec^{2}\Theta }$ - $\dfrac{cosec\Theta }{sec^{2}\Theta }$

To Prove  :

( 1 + cotФ + tanФ ) ( sinФ - cosФ ) = $\dfrac{sec\Theta }{cosec^{2}\Theta }$ - $\dfrac{cosec\Theta }{sec^{2}\Theta }$

Solution :

From Left hand side

( 1 + cotФ + tanФ ) ( sinФ - cosФ ) = $( 1 +\dfrac{sin\Theta }{cos\Theta }+\dfrac{cos\Theta }{sin\Theta })$   ( sinФ - cosФ )

                                                       = $\dfrac{sin\Theta cos\Theta+sin^{2}\Theta +cos^{2}\Theta }{sin\Theta cos\Theta }$ ( sinФ - cosФ )

                                                       = $\dfrac{sin\Theta cos\Theta+1}{sin\Theta cos\Theta }$( sinФ - cosФ )

                                                  = sinФ ( $\dfrac{sin\Theta cos\Theta+1}{sin\Theta cos\Theta }$) - cosФ ( $\dfrac{sin\Theta cos\Theta+1}{sin\Theta cos\Theta }$)    

                                                  = ( $\dfrac{1}{cos\Theta }$ + sinФ ) - (   $\dfrac{1}{sin\Theta }$ + cosФ )  

                                                  = secФ + sinФ - cosecФ - cosФ           .......1

Now,

From Right hand side

 $\dfrac{sec\Theta }{cosec^{2}\Theta }$ - $\dfrac{cosec\Theta }{sec^{2}\Theta }$  =  $\dfrac{\dfrac{1}{cos\Theta }}{\dfrac{1}{sin^{2}\Theta }}$  -     $\dfrac{\dfrac{1}{sin\Theta }}{\dfrac{1}{cos^{2}\Theta }}$            

                             = $\dfrac{sin^{2}\Theta }{cos\Theta }$ -    $\dfrac{cos^{2}\Theta }{sin\Theta }$

                            = $\dfrac{sin^{3}\Theta -cos^{3}\Theta }{cos\Theta sin\Theta }$

                            = $\dfrac{(sin\Theta -cos\Theta )(sin^{2}\Theta +cos^{2}\Theta +sin\Theta cos\Theta )}{sin\Theta cos\Theta }$

                           = $\dfrac{(sin\Theta -cos\Theta )(1 +sin\Theta cos\Theta )}{sin\Theta cos\Theta }$

                           = $\dfrac{1}{cos\Theta }$ + sinФ -   $\dfrac{1}{sin\Theta }$ + cosФ

                          = secФ + sinФ - cosecФ - cosФ         .........2

So From 1 and 2

Left hand side = Right hand side

Hence,  ( 1 + cotФ + tanФ ) ( sinФ - cosФ ) = $\dfrac{sec\Theta }{cosec^{2}\Theta }$ - $\dfrac{cosec\Theta }{sec^{2}\Theta }$  is proved  Answer