Question

prove that ( 1+ cot A + tan A ) ( sin A - cos A ) by sec3 A - cosec 3 A = sin2 A,cos2 A​

Answer 1

Hence, LHS = RHS

Your answer is in attachment.

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Answer 2

Consider LHS

$\sf (cosecA - sinA)(secA - cosA)(cosecA−sinA)</p><p>\sf = \bigg( \dfrac{1}{sinA} - sinA \bigg) \bigg( \dfrac{1}{cosA} - cosA \bigg)$

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

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

[ Because, sin²A + cos²A = 1 ]

$\sf = cosA.sinA=cosA.sinA$

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 ]

$\sf = 1 \times cosA.sinA=1×cosA.sinA$

$\sf = cosA.sinA=cosA.sinA$

LHS = RHS

Hence proved.