Prove that following identities, where the angles involved are acute angles for which the trigonometric ratios as defined: $c o s e c^{6} A - {cot}^{6} A = 3 {cot}^{2} A c o s e c^{2} A + 1$ .

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Solution

$L . H . S . = c o s e c^{6} A - {cot}^{6} A$
= ${(c o s e c^{2} A)}^{3} - {({cot}^{2} A)}^{3}$
= ${(c o s e c^{2} θ - {cot}^{2} A)}^{3} + 3 c o s e c^{2} A {cot}^{2} A (c o s e c^{2} A - {cot}^{2} A)$
= ${(1)}^{3} + 3 c o s e c^{2} A {cot}^{2} A * 1$
= $1 + 3 {cot}^{2} A c o s e c^{2} A$
= $3 {cot}^{2} A c o s e c^{2} A + 1 = R . H . S .$

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