Question

if,
$\tt \dfrac{cos^4\: \alpha}{cos^2\: \beta}+ \dfrac{sin^4\: \alpha}{sin^2\: \beta}=1$
then show that,
$\tt \tt \dfrac{cos^4\: \beta}{cos^2\: \alpha}+ \dfrac{sin^4\: \beta}{sin^2\: \alpha}=1$
​

Answer 1

Step-by-step explanation:

Solution

cos4αcos2β+sin4αsin2β=1

⇒cos4αsin2β+sin4αcos2β=cos2βsin2β

⇒cos4α(1−cos2β)+sin4αcos2β=cos2β(1−cos2β)

⇒cos4α−cos4αcos2β+sin4αcos2β=cos2β−cos4β

⇒cos4α−cos4αcos2β+(1−cos2α)2cos2β=cos2β−cos4β

⇒cos4α−cos4αcos2β+(1+cos4α−2cos2α)2cos2β=cos2β−cos4β

⇒cos4α−cos4αcos2β+cos2β+cos2βcos4α−2cos2αcos2β=cos2β−cos4β

⇒2cos4α=2cos2αcos2β

⇒cos2α=cos2β→(1)

⇒1−sin2α=1−sin2β

⇒sin2α=sin2β→(2)

Now,

(i)L.H.S.=sin4α+sin4β=(sin2α−sinβ)2+2sin2αsin2β

As, sin2α=sin2β,above expression becomes,

=0+2sin2αsin2β=2sin2αsin2β=R.H.S.

(ii)L.H.S.=cos4βcos2α+sin4βsin2α

From (1),

=cos4αcos2α+sin4αsin2α

=cosα+sin2α=1=R.H.S.

Answer 2

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

Given that

$\rm :\longmapsto\:\dfrac{ {cos}^{4} \alpha }{ {cos}^{2}\beta } + \dfrac{ {sin}^{4} \alpha }{ {sin}^{2} \beta } = 1$

On taking LCM, we get

$\rm :\longmapsto\:\dfrac{ {cos}^{4} \alpha {sin}^{2}\beta + {sin}^{4}\alpha {cos}^{2}\beta}{ {cos}^{2}\beta {sin}^{2} \beta } = 1$

can be rewritten as

$\rm :\longmapsto\:\dfrac{ {cos}^{4} \alpha[1 - {cos}^{2}\beta] + {( {sin}^{2}\alpha)}^{2} {cos}^{2}\beta}{ {cos}^{2}\beta[1 - {cos}^{2} \beta]} = 1$

We know,

$\boxed{ \tt{ \: {sin}^{2}x + {cos}^{2}x = 1 \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{ {cos}^{4} \alpha[1 - {cos}^{2}\beta] + {(1 - {cos}^{2}\alpha)}^{2} {cos}^{2}\beta}{ {cos}^{2}\beta[1 - {cos}^{2} \beta]} = 1$

$\rm :\longmapsto\:\dfrac{ {cos}^{4} \alpha[1-{cos}^{2}\beta] + [1 +{cos}^{4}\alpha-{2cos}^{2}\alpha]{cos}^{2}\beta}{ {cos}^{2}\beta[1 - {cos}^{2} \beta]} = 1$

$\rm :\longmapsto\:\dfrac{ {cos}^{4} \alpha- {cos}^{4}\alpha {cos}^{2}\beta + {cos}^{2}\beta + {cos}^{2}\beta {cos}^{4}\alpha-{2cos}^{2}\alpha{cos}^{2}\beta}{ {cos}^{2}\beta[1 - {cos}^{2} \beta]} = 1$

$\rm :\longmapsto\:\dfrac{ {cos}^{4} \alpha + {cos}^{2}\beta -{2cos}^{2}\alpha{cos}^{2}\beta}{ {cos}^{2}\beta[1 - {cos}^{2} \beta]} = 1$

$\rm :\longmapsto\:{cos}^{4} \alpha + {cos}^{2}\beta -{2cos}^{2}\alpha{cos}^{2}\beta = {cos}^{2}\beta - {cos}^{4} \beta$

$\rm :\longmapsto\:{cos}^{4} \alpha -{2cos}^{2}\alpha{cos}^{2}\beta = - {cos}^{4} \beta$

$\rm :\longmapsto\:{cos}^{4} \alpha -{2cos}^{2}\alpha{cos}^{2}\beta + {cos}^{4} \beta = 0$

$\rm :\longmapsto\: {( {cos}^{2}\alpha - {cos}^{2}\beta)}^{2} = 0$

$\rm :\longmapsto\: {cos}^{2}\alpha - {cos}^{2}\beta = 0$

$\red{\rm \implies\: {cos}^{2}\beta = {cos}^{2}\alpha \: }$

Now, Consider

$\rm :\longmapsto\:\dfrac{ {cos}^{4} \beta }{ {cos}^{2} \alpha } + \dfrac{ {sin}^{4}\beta }{ {sin}^{2}\alpha }$

can be rewritten as

$\rm \:  =  \: \dfrac{ {cos}^{4} \beta }{ {cos}^{2} \beta } + \dfrac{ {sin}^{2}\beta \times {sin}^{2} \beta }{ {sin}^{2}\alpha }$

We know,

$\boxed{ \tt{ \: {sin}^{2}x + {cos}^{2}x = 1 \: }}$

So, using this, we get

$\rm \:  =  \: {cos}^{2}\beta+ \dfrac{ {sin}^{2}\beta \times [1 - {cos}^{2} \beta ]}{1 - {cos}^{2}\alpha }$

$\rm \:  =  \: {cos}^{2}\beta+ \dfrac{ {sin}^{2}\beta \times [1 - {cos}^{2} \beta ]}{1 - {cos}^{2} \beta }$

$\rm \:  =  \: {cos}^{2}\beta + {sin}^{2}\beta$

$\rm \:  =  \: 1$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{ {cos}^{4} \beta }{ {cos}^{2} \alpha } + \dfrac{ {sin}^{4}\beta }{ {sin}^{2}\alpha } = 1 \: }}$

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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