Question

prove that
cot³A. sin³A. tan³A. cos³A.
---------------------- + ------------------
(cosA+sinA)² (cosA+sinA)²

= secA. cosecA-1
-----------------------
cosecA+secA​

Answer 1

||✪✪ QUESTION ✪✪||

prove that

[cot³A.sin³A/(cosA+sinA)²] + [tan³A. cos³A / (cosA+sinA)² ]

= (secA. cosecA-1)/(cosecA+secA)

|| ✰✰ ANSWER ✰✰ ||

Solving LHS, First :-

→ [cot³A.sin³A/(cosA+sinA)²] + [tan³A. cos³A / (cosA+sinA)² ]

Putting cotA = cosA/sinA and TanA = sinA/cosA in Numerator , we get,

→ [(cos³A/sin³A).sin³A/(cosA+sinA)²] + [(sin³A/cos³A). cos³A / (cosA+sinA)² ]

→ [ cos³A / (cosA+sinA)²] + [(sin³A / (cosA+sinA)² ]

Taking LCM now, we get,

→ [ ( cos³A + sin³A) / (cosA + sinA)² ]

Now using a³ + b³ = (a+b)(a² + b² - ab) in Numerator ,

→ [ (cosA+sinA)(cos²A+sin²A - cosA*sinA) / (cosA + sinA)² ]

Now, (cosA+sinA) will be cancel ,

→ [ (cos²A+sin²A - cosA*sinA) / (cosA + sinA) ]

Putting cos²A+sin²A = 1 now,

→ [ ( 1 - cosA*sinA) / (cosA + sinA) ] ------------ Equation (1)

______________________________

Now, Solving RHS part :-

→ (secA.cosecA - 1)/(cosecA+secA)

Putting CosecA = 1/sinA and SecA = 1/cosA we get,

→ [ {(1/cosA)*(1/sinA) - 1} / { (1/sinA + (1/cosA)} ]

Taking LCM in both Numerator And Denominator Now, we get,

→ [{(1 - cosA*sinA) / ( sinA*cosA)} / { (cosA + sinA) / (sinA*cosA) } ]

Now, Both Denominator part (sinA*cosA) will be cancel ,

So, we get,

→ [ (1 - cosA*SinA) / (cosA + sinA) ] -------------- Equation (2)

_____________________________

As we can see Now,

☞ Equation (1) = Equation (2)

So,

☛ LHS = RHS .

✪✪ Hence Proved ✪✪

_________________________

Answer 2

QUESTION

prove that,

$\frac{ {cot}^{3} \theta. {sin}^{3} \theta }{ {(cos \theta + sin \theta) }^{2} } + \frac{ {tan}^{3} \theta. {cos}^{3} \theta }{ {(cos \theta + sin \theta ) }^{2} } = \frac{ \sec\theta .cosec \theta - 1 }{cosec \theta + sec \theta }$

Refer to the attachment