Question

trigonometry. ( cosecA - sinA) ( secA - cos A) = 1/ tanA + cotA

Answer 1

☺✌heya mate here is your answer ☺✌


hope this may be helpfull to you ❤


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@alishka ❤

Answer 2

$\huge\bigstar\mathfrak\red{\underline{\underline{SOLUTION:}}}$

Consider the L.H.S,

$(cosec \theta - sin \theta)(sec \theta - cos \theta) \\ = > ( \frac{1}{sin \theta } - sin \theta)( \frac{1}{cos \theta} - cos \theta) \\ \\ = > ( \frac{1 - {sin}^{2} \theta}{sin \theta} )( \frac{ 1 - {cos}^{2} \theta}{cos \theta} ) \\ \\ = > \frac{ {cos}^{2} \theta }{sin \theta} \times \frac{ {sin}^{2} \theta}{cos \theta} = sin \theta \: cos \theta$

Consider the R.H.S,

$\frac{1}{tan \theta + cot \theta} \\ = > \frac{1}{ \frac{sin \theta}{cos \theta} + \frac{cos \theta}{sin \theta} } \\ \\ = > \frac{1}{ \frac{ {sin}^{2} \theta + {cos}^{2} \theta }{sin \theta \:cos \theta} } \\ \\ = > \frac{sin \theta \: cos \theta}{ {cos}^{2} \theta + {sin}^{2} \theta } \\ \\ = > sin \theta \: cos \theta \: \: \: ( {sin}^{2} \theta + {cos}^{2} \theta = 1)$

Since, L.H.S = R.H.S

Therefore,

$= > (cosec \theta - sin \theta)(sec \theta - cos \theta) = \frac{1}{tan \theta + cot \theta}$

hope it helps ☺️