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Basic Trigonometric Identities

Prove that ...

Question

Prove that ${cosec}^{6} θ = {cot}^{6} θ + 3 {cot}^{2} θ . {cosec}^{2} θ + 1$

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Solution

$c o s e c^{6} θ = (c o s e c^{2} θ)^{3} .$

$c o s e c^{2} θ = c o t^{2} θ + 1 - - - - (1)$

$(c o s e c^{2} θ)^{3} = (c o t^{2} θ + 1)^{3}$

Since $(x + y)^{3} = x^{3} + y^{3} + 3 x y (x + y),$ we have

$c o s e c^{6} θ = (c o t^{2} θ)^{3} + (1)^{3} (c o t^{2} θ) (1) (c o t^{2} θ + 1)$

$\Rightarrow c o s e c^{6} θ = c o t^{6} θ + 1 + 3. c o t^{2} . c o s e c^{2} θ$

$\Rightarrow c o s e c^{6} θ = c o t^{6} θ + 3. c o t^{2} c o s e c^{2} θ + 1$

$∴$ L.H.S=R.H.S

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Similar questions

c o s e c^{6} θ - c o t^{6} θ = 1 + 3 c o t^{1} θ + 3 c o t^{4} θ

c o s e c^{6} θ - {cot}^{6} θ = a {cot}^{4} θ + b {cot}^{2} θ + c

a + b + c = ?

{csc}^{6} θ = {cot}^{6} θ + 3 {cot}^{2} θ {csc}^{2} θ + 1

{sec}^{6} θ = {tan}^{6} θ + 3 {tan}^{2} θ {sec}^{2} θ + 1

Q. Prove the following trigonometric identities

cosec⁶θ = cot⁶θ + 3 cot²θ cosec²θ + 1

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