Prove the following identities: 1- + 1- =( cosec x+1)

Question

Prove the following identities: $\frac{\tan\text{x}}{1-\cot\text{x}}+\frac{\cot\text{x}}{1-\tan\text{x}}=(\sec\text{x}\text{ cosec x}+1)$

✓

Answer

$\text{L.H.S}=\frac{\tan\text{x}}{1-\cot\text{x}}+\frac{\cot\text{x}}{1-\tan\text{x}}$ $=\frac{\big(\frac{\sin\text{x}}{\cos\text{x}}\big)}{\big(1-\frac{\cos\text{x}}{\sin\text{x}}\big)}+\frac{\big(\frac{\cos\text{x}}{\sin\text{x}}\big)}{1-\frac{\sin\text{x}}{\cos\text{x}}}$ $=\frac{\sin\text{x}}{\cos\text{x}\frac{\big(\sin\text{x}-\cos\text{x}\big)}{\sin\text{x}}}+\frac{\cos\text{x}}{\sin\text{x}\frac{\big(\cos\text{x}-\sin\text{x}\big)}{\cos\text{x}}}$ $=\frac{\sin^{2}\text{x}}{\cos\text{x}\big(\sin\text{x}-\cos\text{x}\big)}+\frac{\cos^{2}\text{x}}{\sin\text{x}\big(\cos\text{x}-\sin\text{x}\big)}$ $=\frac{\sin^{3}\text{x}-\cos^{3}\text{x}}{\cos\text{x}\sin\text{x}\big(\sin\text{x}-\cos\text{x}\big)}$ $=\frac{\big(\sin\text{x}-\cos\text{x}\big)\big(\sin^{2}\text{x}+\cos^{2}\text{x}+\sin\text{x}\cos\text{x}\big)}{\cos\text{x}\sin\text{x}\big(\sin\text{x}-\cos\text{x}\big)}$ $ \big[\text{using }\text{a}^{3}-\text{b}^{3}=\big(\text{a}-\text{b}\big)\big(\text{a}^{2}+\text{b}^{2}+\text{ab}\big)\big]$ $=\frac{1+\sin\text{x}\cos\text{x}}{\sin\text{x}\cos\text{x}}$ $ \big[\because\sin^{2}\text{x}+\cos^{2}\text{x}=1\big]$ $=\frac{1}{\sin\text{x}\cos\text{x}}+\frac{\sin\text{x}\cos\text{x}}{\sin\text{x}\cos\text{x}}$ $=\sec\text{x}\text{ cosec }\text{x}+1$ $\big[\because\frac{1}{\cos\text{x}}=\sec\text{x},\frac{1}{\sin\text{x}}=\text{cosec }\text{x}\big]$ $=\text{R.H.S}$ $\text{Proved}$