Question

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\bf \: if \: \: \: \frac{ {sin}^{4} \alpha }{a} + \frac{ {cos}^{ 4 } \alpha }{b} = \frac{1}{a + b} \: \: \\ \\ \bf \: then \: prove \: that : \\ \\ \\ \bf \: \frac{ {sin}^{4n} \alpha }{ {a}^{2n - 1} } + \frac{ {cos}^{4n} \alpha }{ {b}^{2n - 1} } = \frac{1}{ {(a + b)}^{2n - 1} }$

​

Answer 1

[

cos

2

α

cosαsinα

cosαsinα

sin

2

α

][

cos

2

β

cosβsinβ

cosβsinβ

sin

2

β

]

=[

cos

2

αcos

2

β+cosαsinαcosβsinβ

cosαsinαcos

2

β+sin

2

αcosβsinβ

cos

2

αcosβsinβ+cosαsinαsin

2

β

cosαsinαcosβsinβ+sin

2

αsin

2

β

]

=[

cosαcosβ(cosαcosβ+sinαsinβ)

sinαcosβ(cosαcosβ+sinαsinβ)

cosαsinβ(cosαcosβ+sinαsinβ)

sinαsinβ(cosαcosβ+sinαsinβ)

]

=[

cosαcosβcos(α−β)

sinαsinβcos(α−β)

cosαcosβcos(α−β)

sinαsinβcos(α−β)

]

Given α−β=(2n+1)

2

π

so, cos(α−β)=0

Thus,[

cosαcosβcos(α−β)

sinαsinβcos(α−β)

cosαcosβcos(α−β)

sinαsinβcos(α−β)

]=[

0

0

0

0

]=0

$[cos2αcosαsinαcosαsinαsin2α][cos2βcosβsinβcosβsinβsin2β]</p><p></p><p></p><p>=[cos2αcos2β+cosαsinαcosβsinβcosαsinαcos2β+sin2αcosβsinβcos2αcosβsinβ+cosαsinαsin2βcosαsinαcosβsinβ+sin2αsin2β]</p><p></p><p></p><p>=[cosαcosβ(cosαcosβ+sinαsinβ)sinαcosβ(cosαcosβ+sinαsinβ)cosαsinβ(cosαcosβ+sinαsinβ)sinαsinβ(cosαcosβ+sinαsinβ)]</p><p></p><p></p><p>=[cosαcosβcos(α−β)sinαsinβcos(α−β)cosαcosβcos(α−β)sinαsinβcos(α−β)]</p><p></p><p></p><p>Given α−β=(2n+1)2π so, cos(α−β)=0</p><p></p><p></p><p>Thus,[cosαcosβcos(α−β)sinαsinβcos(α−β)cosαcosβcos(α−β)sinαsinβcos(α−β)]=[0000]=0</p><p></p><p>$

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Answer By shreyas

Answer 2

Solution−

Given that,

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

$\rm :\longmapsto\: \dfrac{ {sin}^{4} \alpha }{a} + \dfrac{ {( {cos}^{2} \alpha )}^{2} }{b} = \dfrac{1}{a + b}$

$\rm :\longmapsto\: \dfrac{ {sin}^{4} \alpha }{a} + \dfrac{ {(1 - {sin}^{2} \alpha )}^{2} }{b} = \dfrac{1}{a + b}$

$\rm :\longmapsto\: \dfrac{ {sin}^{4} \alpha }{a} + \dfrac{1 + {sin}^{4} \alpha - 2 {sin}^{2} \alpha }{b} = \dfrac{1}{a + b}$

$\rm :\longmapsto\: \dfrac{b {sin}^{4} \alpha + a + a{sin}^{4} \alpha - 2 a{sin}^{2} \alpha }{ab} = \dfrac{1}{a + b}$

$\rm :\longmapsto\: \dfrac{b {sin}^{4} \alpha + a{sin}^{4} \alpha - 2 a{sin}^{2} \alpha }{ab} + \dfrac{a}{ab} = \dfrac{1}{a + b}$

$\rm :\longmapsto\: \dfrac{b {sin}^{4} \alpha + a{sin}^{4} \alpha - 2 a{sin}^{2} \alpha }{ab} + \dfrac{1}{b} = \dfrac{1}{a + b}$

$\rm :\longmapsto\: \dfrac{b {sin}^{4} \alpha + a{sin}^{4} \alpha - 2 a{sin}^{2} \alpha }{ab}= \dfrac{1}{a + b} - \dfrac{1}{b}$

$\rm :\longmapsto\: \dfrac{b {sin}^{4} \alpha + a{sin}^{4} \alpha - 2 a{sin}^{2} \alpha }{ab}= \dfrac{b - a - b}{(a + b)b}$

$\rm :\longmapsto\: \dfrac{b {sin}^{4} \alpha + a{sin}^{4} \alpha - 2 a{sin}^{2} \alpha }{a}= \dfrac{- a}{(a + b)}$

$\rm :\longmapsto\:(a + b) {sin}^{4} \alpha - 2a {sin}^{2} \alpha = - \dfrac{ {a}^{2} }{a + b}$

$\rm :\longmapsto\:(a + b)^{2} {sin}^{4} \alpha - 2a(a + b) {sin}^{2} \alpha = - {a}^{2}$

$\rm :\longmapsto\:(a + b)^{2} {sin}^{4} \alpha - 2a(a + b) {sin}^{2} \alpha + {a}^{2} = 0$

$\bf\implies \: {\bigg[(a + b) {sin}^{2} \alpha - a \bigg]}^{2} = 0$

$\bf\implies \: {\bigg[(a + b) {sin}^{2} \alpha - a \bigg]} = 0$

$\bf\implies \: {sin}^{2} \alpha = \dfrac{a}{a + b}$

So,

$\bf\implies \: {cos}^{2} \alpha = 1 - {sin}^{2} \alpha = 1 - \dfrac{a}{a + b} = \dfrac{b}{a + b}$

Now, Consider

$\rm :\longmapsto\:\dfrac{ {sin}^{4n} \alpha }{ {a}^{2n - 1} } + \dfrac{ {cos}^{4n} \alpha }{ {b}^{2n - 1} }$

$\rm \: = \: \dfrac{ ({sin}^{2} \alpha) {}^{2n} }{ {a}^{2n - 1} } + \dfrac{ ({cos}^{2} \alpha)^{2n} }{ {b}^{2n - 1} }$

$\rm \: = \: \dfrac{ {a}^{2n} }{ {(a + b)}^{2n} {a}^{2n - 1} } + \dfrac{ {b}^{2n} }{ {(a + b)}^{2n} {b}^{2n - 1} }$

$\rm \: = \: \dfrac{ {a}^{} }{ {(a + b)}^{2n}} + \dfrac{ {b}^{} }{ {(a + b)}^{2n} }$

$\rm \: = \: \dfrac{ {a} + b}{ {(a + b)}^{2n}}$

$\rm \: = \: \dfrac{1}{ {(a + b)}^{2n - 1}}$

Hence, Proved