Question

Cosec6A-cot6A=3cot2Acosec2A+1

Answer 1

Proved

•LHS

•cosec^6A-cot^6A

•(cosec²A)³- (cot²A)³

•Using a³- b³ = ( a - b )( a² + b² + ab )

•(cosec²A-cot²A){(cosec²A)²+

(cot²A)²+cosec²Acot²A}

____[ By Using, cosec²A= 1+cot²A

cosec²A-cot²A =1]

•{(cosec²A)²+(cot²A)²+cosec²Acot²A}

•Now adding and subtracting

2cosec²Acot²A

•(cosec²A)²+(cot²A)²-2cosec²Acot²A + 3cosec²Acot²A

•{(cosec²A-cot²A)² + 3cosec²Acot²A}

•1 +3cosec²Acot²A

•RHS

Answer 2

$\csc^6A-\cot^6A=3\cot^2A\csc^2A+1$, proved.

Step-by-step explanation:

To prove that:

$\csc^6A-\cot^6A=3\cot^2A\csc^2A+1$

L.H.S. = $\csc^6A-\cot^6A$

= $(\csc^2A)^3-(\cot^2A)^3$

Using the algebraic identity,

$(a-b)^{3} =a^3-b^3-3ab(a-b)$

⇒ $a^3-b^3=(a-b)^{3} +3ab(a-b)$

= $(\csc^2A-\cot^2A)^3+3\csc^2A\cot^2A(\csc^2A-\cot^2A)$

Using the trigonometric identity,

$\csc^2\theta-\cot^2\theta$ = 1

= $(1)^3+3\csc^2A\cot^2A(1)$

= 1 + $3\csc^2A\cot^2A$

= $3\cot^2A\csc^2A$ + 1

= R.H.S., proved.

Thus, $\csc^6A-\cot^6A=3\cot^2A\csc^2A+1$, proved.