Question

Prove that:
Sec^3A.Cosec^3A-3SecA.CosecA=Tan^3+Cot^3​

Answer 1

Appropriate Question :-

Prove that

$\rm \: \: {sec}^{3}A \: {cosec}^{3}A - 3 \: secA \: cosecA = {tan}^{3}A + {cot}^{3} A$

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

Consider RHS

$\rm :\longmapsto\: {tan}^{3}A + {cot}^{3}A$

can be rewritten as

$\rm \: = \: {(tanA)}^{3} + {(cotA)}^{3}$

We know,

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

So, using this identity, we get

$\rm \: = \: {(tanA + cotA)}^{3} - 3tanA \: cotA \: (tanA + cotA)$

$\rm = {\bigg[\dfrac{sinA}{cosA} + \dfrac{cosA}{sinA} \bigg]}^{3} - 3tanA \times \bigg[\dfrac{1}{tanA} \bigg]\bigg[\dfrac{sinA}{cosA} + \dfrac{cosA}{sinA} \bigg]$

$\rm = {\bigg[\dfrac{ {sin}^{2} A + {cos}^{2}A}{sinA \: cosA} \bigg]}^{3} - 3\bigg[\dfrac{ {sin}^{2} A + {cos}^{2}A}{sinA \: cosA} \bigg]$

$\rm = {\bigg[\dfrac{ 1}{sinA \: cosA} \bigg]}^{3} - 3\bigg[\dfrac{ 1}{sinA \: cosA} \bigg]$

We know,

$\boxed{ \tt{ \: \frac{1}{sinx} = cosecx}} \: \: \: and \: \: \: \boxed{ \tt{ \: \frac{1}{cosx} = secx}}$

So, using this, we get

$\rm \: = \: {\bigg[secAcosecA\bigg]}^{3} - 3secAcosecA$

$\rm \: = \: {sec}^{3}A \: {cosec}^{3}A - 3 \: secA \: cosecA$

Hence,

$\boxed{ \tt{{sec}^{3}A{cosec}^{3}A - 3secAcosecA={tan}^{3}A+{cot}^{3} A}}$

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

Answer:

Appropriate Question :-

Prove that

$\rm \: \: {sec}^{3}A \: {cosec}^{3}A - 3 \: secA \: cosecA = {tan}^{3}A + {cot}^{3} A$

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

Consider RHS

$\rm :\longmapsto\: {tan}^{3}A + {cot}^{3}A$

can be rewritten as

$\rm \: = \: {(tanA)}^{3} + {(cotA)}^{3}$

We know,

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

So, using this identity, we get

$\rm \: = \: {(tanA + cotA)}^{3} - 3tanA \: cotA \: (tanA + cotA)$

$\rm = {\bigg[\dfrac{sinA}{cosA} + \dfrac{cosA}{sinA} \bigg]}^{3} - 3tanA \times \bigg[\dfrac{1}{tanA} \bigg]\bigg[\dfrac{sinA}{cosA} + \dfrac{cosA}{sinA} \bigg]$

$\rm = {\bigg[\dfrac{ {sin}^{2} A + {cos}^{2}A}{sinA \: cosA} \bigg]}^{3} - 3\bigg[\dfrac{ {sin}^{2} A + {cos}^{2}A}{sinA \: cosA} \bigg]$

$\rm = {\bigg[\dfrac{ 1}{sinA \: cosA} \bigg]}^{3} - 3\bigg[\dfrac{ 1}{sinA \: cosA} \bigg]$

We know,

$\boxed{ \tt{ \: \frac{1}{sinx} = cosecx}} \: \: \: and \: \: \: \boxed{ \tt{ \: \frac{1}{cosx} = secx}}$

So, using this, we get

$\rm \: = \: {\bigg[secAcosecA\bigg]}^{3} - 3secAcosecA$

$\rm \: = \: {sec}^{3}A \: {cosec}^{3}A - 3 \: secA \: cosecA$

Hence,

$\boxed{ \tt{{sec}^{3}A{cosec}^{3}A - 3secAcosecA={tan}^{3}A+{cot}^{3} A}}$

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