Krishna
0

Step 1:  Note down the given equation

              \left(\operatorname{cosec}\theta-\cot\theta\right)^2=\frac{1-\cos\theta}{1+\cos\theta}


Step 2: Manipulate the L.H.S by the equivalent trigonometric ratios

            L.H.S:  = \left(\operatorname{cosec}\theta-\cot\theta\right)^2

                        =    (\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta})^2


                        Do the L.C.M


                       =   (\frac{1 - \cos \theta}{\sin \theta})^2


                       =   (\frac{(1 - \cos \theta)^2}{\sin^2 \theta}


                        =   (\frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}

                                                [since, \sin^2 \theta + \cos^2 \theta = 1


Step 3: Use the a^2 - b^2 = (a - b) (a - b) formula to simplify

              EXAMPLE: = \frac{(1-\cos\theta)^2}{\left(1^2-\cos^2\theta\right)}


                                =  \frac{(1-\cos\theta)^2}{(1+\cos\theta)(1-\cos\theta)}


                              Cancel the equal terms

                                = \frac{(1-\cos\theta)}{(1+\cos\theta)}

                                      

                              Hence, proved R.H.S