Question

sec^6theta- tan^6theta = 3 tan^2 theta sec^2theta +1​

Answer 1

Question :-

Prove that,

$\sf \: \: {sec}^{6}\theta - {tan}^{6}\theta =  \: 1 + 3 {sec}^{2}\theta {tan}^{2}\theta$

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

Consider,

$\rm :\longmapsto\: {sec}^{6}\theta - {tan}^{6}\theta$

can be rewritten as

$\rm \:  =  \: {( {sec}^{2}\theta ) }^{3} - {( {tan}^{2}\theta ) }^{3}$

We know,

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

So, using this identity, we get

$\rm \:  =  \: {\bigg[ {sec}^{2}\theta - {tan}^{2}\theta\bigg]}^{3} + 3 {sec}^{2}\theta {tan}^{2}\theta \bigg[ {sec}^{2}\theta-{tan}^{2}\theta\bigg]$

We know,

$\boxed{ \bf{ \: {sec}^{2}\theta - {tan}^{2}\theta = 1}}$

So, on substituting the values, we get

$\rm \:  =  \: {(1)}^{3} + 3 {sec}^{2}\theta {tan}^{2}\theta (1)$

$\rm \:  =  \: 1 + 3 {sec}^{2}\theta {tan}^{2}\theta$

Hence,

$\rm \: \boxed{ \bf{ \: {sec}^{6}\theta - {tan}^{6}\theta =  \: 1 + 3 {sec}^{2}\theta {tan}^{2}\theta }}$

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

Answer 2

Question :-

Prove that,

$\sf \: \: {sec}^{6}\theta - {tan}^{6}\theta =  \: 1 + 3 {sec}^{2}\theta {tan}^{2}\theta$

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

Consider,

$\rm :\longmapsto\: {sec}^{6}\theta - {tan}^{6}\theta$

can be rewritten as

$\rm \:  =  \: {( {sec}^{2}\theta ) }^{3} - {( {tan}^{2}\theta ) }^{3}$

We know,

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

So, using this identity, we get

$\rm \:  =  \: {\bigg[ {sec}^{2}\theta - {tan}^{2}\theta\bigg]}^{3} + 3 {sec}^{2}\theta {tan}^{2}\theta \bigg[ {sec}^{2}\theta-{tan}^{2}\theta\bigg]$

We know,

$\boxed{ \bf{ \: {sec}^{2}\theta - {tan}^{2}\theta = 1}}$

So, on substituting the values, we get

$\rm \:  =  \: {(1)}^{3} + 3 {sec}^{2}\theta {tan}^{2}\theta (1)$

$\rm \:  =  \: 1 + 3 {sec}^{2}\theta {tan}^{2}\theta$

Hence,

$\rm \: \boxed{ \bf{ \: {sec}^{6}\theta - {tan}^{6}\theta =  \: 1 + 3 {sec}^{2}\theta {tan}^{2}\theta }}$

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