Question

if cos to the power 4 Alpha divided by cos squared theta + sin to the power 4 alpha by sin square theta equals to 1 then prove that cos to the power 4 theta by cos square alpha + sin to the power 4 theta by sin square alpha equals to 1​

Answer 1

To prove: $\frac{cos^{4}\beta}{cos^{2}\alpha }+\frac{sin^{4}\beta}{sin^{2}\alpha}=1$

Given

$\frac{cos^{4}\alpha}{cos^{2}\beta}+\frac{sin^{4}\alpha}{sin^{2}\beta }=1$

$\frac{cos^{4}\alpha sin^{2}\beta+sin^{4}\alpha cos^{2}\beta}{cos^{2}\beta sin^{2}\beta}=1$

$cos^{4}\alpha sin^{2}\beta+sin^{4}\alpha cos^{2}\beta}=cos^{2}\beta sin^{2}\beta$

$cos^{4}\alpha sin^{2}\beta+sin^{4}\alpha (1-sin^{2}\beta)=(1-sin^{2}\beta)sin^{2}\beta$

                        [ ∵$cos^{2}\beta=1-sin^{2}\beta$  ]

$cos^{4}\alpha sin^{2}\beta+sin^{4}\alpha-sin^{4}\alpha sin^{2}\beta=sin^{2}\beta-sin^{4}\beta$

$(1-sin^{2}\alpha)^{2} sin^{2}\beta+sin^{4}\alpha-sin^{4}\alpha sin^{2}\beta=sin^{2}\beta-sin^{4}\beta$

                            [ ∵$cos^{2}\alpha =1-sin^{2}\alpha$ ]

$(1+sin^{4}\alpha-2sin^{2}\alpha } )sin^{2}\beta+sin^{4}\alpha-sin^{4}\alpha sin^{2}\beta=sin^{2}\beta-sin^{4}\beta$

$sin^{2}\beta+sin^{4}\alpha sin^{2}\beta-2sin^{2}\alpha sin^{2}\beta+} sin^{4}\alpha-sin^{4}\alpha sin^{2}\beta=sin^{2}\beta-sin^{4}\beta$

$sin^{4}\beta-2sin^{2}\alpha sin^{2}\beta+ sin^{4}\alpha=0$

$(sin^{2}\beta-sin^{2}\alpha )^{2} =0$

$sin^{2}\beta=sin^{2}\alpha---------(1)$

$1-cos^{2}\beta=1-cos^{2}\alpha$

$cos^{2}\beta=cos^{2}\alpha----------(2)$

Consider,

$\frac{cos^{4}\beta}{cos^{2}\alpha }+\frac{sin^{4}\beta}{sin^{2}\alpha}=\frac{cos^{2}\beta cos^{2}\beta}{cos^{2}\beta }+\frac{sin^{2}\beta sin^{2}\beta}{sin^{2}\beta }$        [from (1) and (2)]

$\frac{cos^{4}\beta}{cos^{2}\alpha }+\frac{sin^{4}\beta}{sin^{2}\alpha}=cos^{2}\beta+sin^{2}\beta$

$\frac{cos^{4}\beta}{cos^{2}\alpha }+\frac{sin^{4}\beta}{sin^{2}\alpha}=1$

Hence proved

Note: I used beta in place of theta as there was no option in equations for theta.

Answer 2

Cos⁴θ /Cos²α  + Sin⁴θ/Sin²α  = 1  if Cos⁴α /Cos²θ  + Sin⁴α/Sin²θ  = 1

Step-by-step explanation:

Given

Cos⁴α /Cos²θ  + Sin⁴α/Sin²θ  = 1

to be proved

Cos⁴θ /Cos²α  + Sin⁴θ/Sin²α  = 1

Cos⁴α /Cos²θ  + Sin⁴α/Sin²θ  = 1

=> Cos⁴αSin²θ  + Sin⁴αCos²θ = Cos²θSin²θ

=> Cos⁴α(1 - Cos²θ)  + (1 - Cos²α)²Cos²θ = Cos²θ(1 - Cos²θ)

=>  Cos⁴α - Cos⁴αCos²θ  + (1  + Cos⁴α - 2Cos²α)Cos²θ = Cos²θ - Cos⁴θ

=>Cos⁴α - Cos⁴αCos²θ  + Cos²θ  + Cos⁴αCos²θ - 2Cos²αCos²θ = Cos²θ - Cos⁴θ

=> Cos⁴α  - 2Cos²αCos²θ + Cos⁴θ  = 0

=> (Cos²α - Cos²θ)² = 0

=> Cos²α - Cos²θ = 0

=>  Cos²α = Cos²θ

LHS

Cos⁴θ /Cos²α  + Sin⁴θ/Sin²α

= (Cos²α)²/Cos²α  + (1 - Cos²θ)²/(1 -Cos²α)

= Cos²α + (1 - Cos²α)²/(1 -Cos²α)

=  Cos²α + 1 - Cos²α

= 1

= RHS

QED

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