Question

Prove the following identities (1-16)

$\frac{\left(1+\cot x+\tan x\right)\left(\sin x-\cos x\right)}{{\sec }^{3}x-{\mathrm{cosec}}^{3}x}={\sin }^{2}x{\cos }^{2}x$

Answer 1

$\frac{\left(1+\cot x+\tan x\right)\left(\sin x-\cos x\right)}{{\sec }^{3}x-{\mathrm{cosec}}^{3}x}={\sin }^{2}x{\cos }^{2}x$

$$\begin{gathered} \mathrm{LHS}=\frac{\left(1+\cot x+\tan x\right)\left(\sin x-\cos x\right)}{{\sec }^{3}x-{\mathrm{cosec}}^{3}x} \\ =\frac{\left(1+{\displaystyle \frac{\cos x}{\sin x}}+{\displaystyle \frac{\sin x}{\cos x}}\right)\left(\sin x-\cos x\right)}{{\displaystyle \frac{1}{{\cos }^{3}x}}-{\displaystyle \frac{1}{{\sin }^{3}x}}} \\ =\frac{\left(\sin x\cos x+{\cos }^{2}x+{\sin }^{2}x\right)\left(\sin x-\cos x\right)\left({\sin }^{2}x{\cos }^{2}x\right)}{\left({\sin }^{3}x-{\cos }^{3}x\right)} \\ =\frac{\left(1+\sin x\cos x\right)\left(\sin x-\cos x\right)\left({\sin }^{2}x{\cos }^{2}x\right)}{\left(\sin x-\cos x\right)\left({\sin }^{2}x+{\cos }^{2}x+\sin x\cos x\right)} \\ ={\sin }^{2}x{\cos }^{2}x \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$