Question

(cosecx - sinx) ( secx - cosx ) = 1 /tanx + cotx​

Answer 1

Step-by-step explanation:

1/sin - sin = 1- sin^2 /sin

1-cos^2 / cos

cos^2 /sin × sin^2 /cos

= single

Answer 2

$\bold{\red {Question}}$

$Prove \: that \: (cosec \: x - sin \: x) \: (sec \: x - cos \: x) \: = \: 1/tanx \: - \: cotx?</p><p></p><p>$

$\huge \:Solution$

$\pink{RHS} \: should \: be \: 1/(tanx \: + \: cotx) \: or \: you \: wont \: get \: the \: answer</p><p></p><p> \: (cosec \: x - sin \: x) \: (sec \: x - cos \: x) \: = \: 1 / \: (tan \: x + cot \: x) \: </p><p> \: (cosec \: x - sin \: x) \: (sec \: x - cos \: x) \: (tan \: x \: + cot \: x) \: = \: 1 \: \: \: \: \: \: \: \: \: \: \: \: \:</p><p></p><p> \green{LHS } \: = \: (cosec \: x - sin \: x) \: (sec \: x - cos \: x) \: (tan \: x + cot \: x)</p><p> \: = \: (1 / sin \: x - sin \: x) \: (1 / cos \: x - cos \: x) \: (tan \: x + 1 / tan \: x)</p><p> \: = \: (1 - sin²x) \: (1 - cos²x)(tan²x + 1) \: / \: (sin \: x * cos \: x * tan \: x)</p><p> \: = \: cos²x * sin²x * \: (tan²x \: + \: 1) \: / (sin \: x * cos \: x * tan \: x), \: \red{ noting \: sin²x + cos²x = 1}</p><p>= \: cos²x \: * \: sin²x \: * sec²x \: / [sin \: x * cos \: x * (sin \: x / cos \: x)], \pink {noting tan²x + 1 = sec²x}</p><p> \: = \: sin²x / sin²x, \: as \: sec²x \: = \: 1 / cos²x</p><p> \: = 1 \: </p><p> \: = \: RHS$