Question

cosec⁶ theta-cot⁶ theta =1 + 3 cosec² theta. cot²theta
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Answer 1

$\large\underline{\sf{Solution-}}$

Consider, LHS

$\rm :\longmapsto\: {cosec}^{6}\theta - {cot}^{6}\theta$

can be rewritten as

$\rm \:  =  \: {( {cosec}^{2} \theta )}^{3} - {( {cot}^{2} \theta )}^{3}$

We know,

$\boxed{\tt{ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y)}}$

So, using this identity, we get

$\rm \:  =  \: {( {cosec}^{2} \theta - {cot}^{2} \theta )}^{3} + 3 {cosec}^{2}\theta {cot}^{2}\theta ( {cosec}^{2}\theta - {cot}^{2}\theta )$

We know,

$\boxed{\tt{ {cosec}^{2}x - {cot}^{2}x = 1 \: }}$

So, using this identity, we get

$\rm \:  =  \: {(1)}^{3} + 3 {cosec}^{2}\theta {cot}^{2}\theta (1)$

$\rm \:  =  \: 1 + 3 {cosec}^{2}\theta {cot}^{2}\theta$

Hence,

$\rm\implies \: \boxed{\tt{ {cosec}^{6}\theta - {cot}^{6}\theta = 3 {cosec}^{2}\theta {cot}^{2}\theta }}$

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

Answer 2

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Consider, LHS

$\rm :\longmapsto\: {cosec}^{6}\theta - {cot}^{6}\theta$

can be rewritten as

$\rm \:  =  \: {( {cosec}^{2} \theta )}^{3} - {( {cot}^{2} \theta )}^{3}$

We know,

$\boxed{\tt{ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y)}}$

So, using this identity, we get

$\rm \:  =  \: {( {cosec}^{2} \theta - {cot}^{2} \theta )}^{3} + 3 {cosec}^{2}\theta {cot}^{2}\theta ( {cosec}^{2}\theta - {cot}^{2}\theta )$

We know,

$\boxed{\tt{ {cosec}^{2}x - {cot}^{2}x = 1 \: }}$

So, using this identity, we get

$\rm \:  =  \: {(1)}^{3} + 3 {cosec}^{2}\theta {cot}^{2}\theta (1)$

$\rm \:  =  \: 1 + 3 {cosec}^{2}\theta {cot}^{2}\theta$

Hence,

$\rm\implies \: \boxed{\tt{ {cosec}^{6}\theta - {cot}^{6}\theta = 3 {cosec}^{2}\theta {cot}^{2}\theta }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1