Saved Bookmarks
| 1. |
Prove that 1/ cosec x + cot x – 1/sin x = 1/sin x –1/ cosec x–cot x |
|
Answer» In order to show that, \(\frac1{(coses\,x+cot\,x)}-\frac1{sin\,x}=\frac1{sin\,x}-\frac1{(coses\,x-cot\,x)}\) it is sufficient to show \(\frac1{coses\,x+cot\,x}+\frac1{(coses\,x-cot\,x)}=\frac1{sin\,x}+\frac1{sin\,x}\) \(\Rightarrow\frac1{coses\,x+cot\,x}+\frac1{(coses\,x-cot\,x)}=\frac2{sin\,x}\;\;...(i)\) Now, LHS of above is \(\frac1{(coses\,x+cot\,x)}+\frac1{(coses\,x-cot\,x)}\) \(=\frac{(coses\,x-cot\,x)+(coses\,x+cot\,x)}{(coses\,x-cot\,x)(coses\,x+cot\,x)}\) \(=\frac{2\,coses\,x}{coses^2x-cot^2x}\) [∵ (a+b)(a-b) = a2 - b2) \(=\frac{2\,coses\,x}1=\frac2{sin\,x}\) = RHS of (i) Hence, \(\frac1{(coses\,x+cot\,x)}+\frac1{(coses\,x-cot\,x)}=\frac1{sin\,x}+\frac1{sin\,x}\) or \(\frac1{(coses\,x+cot\,x)}-\frac1{sin\,x}=\frac1{sin\,x}-\frac1{(coses\,x-cot\,x)}.\) |
|