Question

prove that sin@/1-cot@+cos@/1-tan@=cos@+sin@​

Answer 1

Answer:

your proof is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: prove \: that : \\ \frac{sin \: x}{1 - cot \: x} + \frac{cos \: x}{1 - tan \: x} = cos \: x + sin \: x \\ \\ LHS = \frac{sin \: x}{1 - cot \: x} + \frac{cos \: x}{1 - tan \: x} \\ \\ = \frac{sin \: x}{1 - cot \: x} + \frac{cos \: x}{1 - \frac{1}{cot \: x} } \\ \\ = \frac{sin \: x}{1 - cot \: x} + \frac{cos \: x.cot \: x}{cot \: x - 1} \\ \\ = \frac{sin \: x}{1 - cot \: x} + \frac{cosx.cot \: x}{ - (1 - cot \: x)} \\ \\ = \frac{sin \: x}{1 - cot \: x} - \frac{cos \: x.cot \: x}{1 - cot \: x} \\ \\ = \frac{sin \: x - cos \: x.cot \: x}{1 - cot \: x} \\ \\ = \frac{sin \: x - cos \: x. \frac{cos \: x}{sin \: x} }{1 - \frac{cos \: x}{sin \: x} } \\ \\ = \frac{ \frac{ \frac{sin {}^{2} \: x - cos {}^{2} \: x }{sin \: x} }{sin \: x - cos \: x} }{sin \: x} \\ \\ = \frac{sin {}^{2} \: x - cos {}^{2} \: x }{sin \: x - cos \: x} \\ \\ = \frac{(sin \: x + cos \: x)(sin \: x - cos \: x)}{sin \: x - cos \: x} \\ \\ = sin \: x + cos \: x \\ \\ = RHS$