Question

prove that (cosec A - 3 sin A)(sec A - cos A ) =1/tan A + cot A​

Answer 1

Correct Question :-

Prove that (cosec A - sin A)(sec A - cos A ) =1/(tan A + cot A)

Solution :-

$\sf (cosecA - sinA)(secA - cosA) = \dfrac{1}{tan A + cotA}$

Consider LHS

$\sf (cosecA - sinA)(secA - cosA)$

$\sf = \bigg( \dfrac{1}{sinA} - sinA \bigg) \bigg( \dfrac{1}{cosA} - cosA \bigg)$

Taking LCM

$\sf = \bigg( \dfrac{1 - sin ^{2} A }{sinA} \bigg) \bigg( \dfrac{1 - cos^{2} A }{cosA} \bigg)$

$\sf = \bigg( \dfrac{cos ^{2} A }{sinA} \bigg) \bigg( \dfrac{sin^{2} A }{cosA} \bigg)$

[ Because, sin²A + cos²A = 1 ]

$\sf = cosA.sinA$

Consider RHS

$\sf \dfrac{1}{tan A + cotA}$

$\displaystyle{ \sf = \frac{1}{ \dfrac{sinA}{cosA} + \dfrac{cosA}{sinA} }}$

[ Because tanA = sinA/cosA and cotA = cosA/sinA ]

Taking LCM

$\sf = \dfrac{1}{ \frac{sin^{2} A+ cos^{2}A} {cosA.sinA}}$

$\sf = \dfrac{1}{ \frac{1} {cosA.sinA}}$

[ Because sin²A + cos²A = 1 ]

$\sf = 1 \times cosA.sinA$

$\sf = cosA.sinA$

LHS = RHS

Hence proved.