Question

If cos⁴alpha by cos²beta +sin⁴alpha by sin² beta =1 prove that cos⁴ beta by cos²alpha + sin⁴beta by sin² alpha =1

Answer 1

Answer:

Step-by-step explanation:

Answer 2

From the given equation we have proved that $\frac{cos^4\beta}{cos^2\alpha}+\frac{sin^4\beta}{sin^2\alpha}=1$  

Hence proved Step-by-step explanation: Given equation is $\frac{cos^4\alpha}{cos^2\beta}+\frac{sin^4\alpha}{sin^2\beta}=1$

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

Now to solve the given equation :

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

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

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

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

here a=1 and $b=sin^2\alpha$ )

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

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

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

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

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

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

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

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

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

Rewritting the above equation we have

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

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

( by using the identity $(a-b)^2=a^2-2ab+b^2$ here $a=sin^2\alpha$ and $b=sin^2\beta$ )

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

$sin^2\alpha-sin^2\beta=0$

$sin^2\alpha=sin^2\beta$

Which implies that $2\alpha=2\beta$

Therefore $\alpha=\beta$

Now we have to prove that $\frac{cos^4\beta}{cos^2\alpha}+\frac{sin^4\beta}{sin^2\alpha}=1$

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

The above expression  becomes

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

$=\frac{(cos^2\alpha)^2}{cos^2\alpha}+\frac{(sin^2\alpha)^2}{sin^2\alpha}$  

$=cos^2\alpha+sin^2\alpha$  

( by the identity $cos^\theta+sin^2\theta=1$  here $\theta=\alpha$  )

Therefore LHS=RHS

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

Hence proved