Question

prove that
tan theta - cot theta / sin theta cos theta = tan²theta - cot²theta​

Answer 1

Step-by-step explanation:

the answer has been attached

Answer 2

Answer:

                                 $\frac{tan theta - cot theta }{sin theta.cos theta}$ = $tan^{2}$Ф - $cot^{2}$ Ф

L.H.S = $\frac{tan theta - cot theta}{sin theta.cos theta}$

         = $\frac{tan theta \frac{1}{tan theta} }{sin theta.cos theta}$

        = $\frac{tan^{2} theta - 1}{tan theta.sin theta.cos theta}$

        = $\frac{tan^{2} theta - 1 }{tan theta.\frac{sin theta}{cos theta}.cos theta. cos theta}$

        =  $\frac{(tan^{2} theta - 1) sec^{2} theta }{tan^{2} theta}$  

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

        = $\frac{tan^{4} theta - 1}{tan^{2} theta}$  

       = $\frac{tan^{4} theta }{tan^{2} theta } - \frac{1}{tan^{2} theta }$

       = $tan^{2}$Ф - $cot^{2}$Ф = R.H.S

Hope this helps you!!