Question

SOLVE THIS QUESTION ​

Answer 1

• According To Question ; $\sf{(\cos^4\alpha/\cos^2\beta) + (\sin^4\alpha/\sin^2\beta) = 1}$

The expression comprised of factors that appear either in $\sf{\cos^2\beta }$ and $\sf{\sin^2\beta }$ is the Least Common Multiplier of $\sf{\cos^2\beta }$ and $\sf{\sin^2\beta }$ .  Hence we can say $\sf{\cos^2\beta \sin^2\beta}$ is the LCM (Least Common Multiplier) of $\sf{\cos^2\beta }$ and $\sf{\sin^2\beta }$

$\longrightarrow\sf{[\cos ^{4} \alpha\left(\sin ^{2} \beta\right)+\sin ^{4} \alpha\left(\cos ^{2} \beta\right)]/[\cos ^{2} \beta \cdot \sin ^{2} \beta] = 1}$

Multiply Both Sides of the Equation by $\sf{\cos ^{2} \beta \cdot \sin ^{2} \beta}$

$\longrightarrow\sf{\cos ^{4} \alpha\left(\sin ^{2} \beta\right)+\sin ^{4} \alpha\left(\cos ^{2} \beta\right)= \cos ^{2} \beta \cdot \sin ^{2} \beta}$

We know $\sf{\sin^2\theta+\cos^2\theta = 1}$ . From $\sf{\sin^2\theta+\cos^2\theta = 1}$ ; $\sf{\sin^2\theta = 1-\cos^2\theta}$

$\longrightarrow\sf{\cos ^{4} \alpha\left(1-\cos^2\beta\right)+\sin ^{4} \alpha\left(\cos ^{2} \beta\right)= \cos ^{2} \beta \cdot (1-\cos^2\beta)}$

Expand the Multiplications : $\sf{a(b-c) = ab - ac}$

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

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

We know $\sf{\sin^2\theta+\cos^2\theta = 1}$ . From $\sf{\sin^2\theta+\cos^2\theta = 1}$ ; $\sf{\sin^2\theta = 1-\cos^2\theta}$

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

Expand $\sf{(1-\cos\alpha)^2}$ by applying $\sf{(a+b)^2 = a^2+b^2+2ab}$ identity

$\longrightarrow\sf{\cos ^{4} \alpha-\cos ^{4} \alpha \cos ^{2} \beta+\left(1-2 \cos ^{2} \alpha+\cos ^{4} \alpha\right) \cos ^{2} \beta=\cos ^{2} \beta-\cos ^{4} \beta}$

Expand the Multiplications : $\sf{a(b-c + d) = ab - ac + ad}$

$\to\sf{\cos ^{4} \alpha-\cos ^{4} \alpha \cos ^{2} \beta+\left(\cos ^{2} \beta-2\cos ^{2} \beta \cos ^{2} \alpha+\cos ^{2} \beta\cos ^{4} \alpha\right) =\cos ^{2} \beta-\cos ^{4} \beta}$$\to\sf{\cos ^{4} \alpha-\cos ^{4} \alpha \cos ^{2} \beta+\cos ^{2} \beta-2\cos ^{2} \beta \cos ^{2} \alpha+\cos ^{2} \beta\cos ^{4} \alpha =\cos ^{2} \beta-\cos ^{4} \beta}$

Rearrange the Like Terms which helps in Easy Calculations

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

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

Subtract $\sf{\cos ^{2} \beta}$ from both Sides of the Equation

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

$\longrightarrow\sf{\cos ^{4} \alpha -2\cos ^{2} \beta \cos ^{2} \alpha =-\cos ^{4} \beta}$

Adding $\sf{\cos ^{4} \beta}$ to both Sides of the Equation

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

$\longrightarrow\sf{\cos ^{4} \alpha -2\cos ^{2} \beta \cos ^{2} \alpha + \cos ^{4} \beta =0}\longrightarrow\sf{\left(\cos ^{2} \alpha-\cos ^{2} \beta\right)^{2}=0}$

Take Square Root on Both Sides of the Equation

$\longrightarrow\sf{\sqrt{\left(\cos ^{2} \alpha-\cos ^{2} \beta\right)^{2}}=0}\longrightarrow\sf{\cos ^{2} \alpha-\cos ^{2} \beta=0}$

Adding $\sf{\cos ^{2} \beta}$ to both Sides of the Equation

$\longrightarrow\sf{\cos ^{2} \alpha-\cos ^{2} \beta+\cos ^{2} \beta=0+\cos ^{2} \beta}\longrightarrow\sf{\boxed{\cos ^{2} \alpha=\cos ^{2} \beta}}$

We know $\sf{\sin^2\theta+\cos^2\theta = 1}$ . From $\sf{\sin^2\theta+\cos^2\theta = 1}$ ; $\sf{\cos^2\theta = 1-\sin^2\theta}$

$\longrightarrow\sf{1-\sin ^{2} \alpha=1-\sin ^{2} \beta}\longrightarrow\sf{-\sin ^{2} \alpha = -\sin ^{2} \beta }\longrightarrow\sf{\boxed{\sin ^{2} \alpha = \sin ^{2} \beta}}$

• We need to Prove that ; $\sf{(\cos^4\beta /\cos^2\alpha) + (\sin^4\beta/\sin^2\alpha ) = 1}$

Since $\sf{\cos^2\alpha =\cos^2\beta }$ and $\sf{\sin^2\alpha =\sin^2\beta }$ from above Calculations

$\longrightarrow\sf{(\cos^4\beta /\cos^2\beta) + (\sin^4\beta/\sin^2\beta ) = 1}$

$\longrightarrow\sf{\cos^{4-2}\beta+ \sin^{4-2}\beta = 1}\longrightarrow\sf{\cos^{2}\beta+ \sin^{2}\beta = 1}\longrightarrow\sf{1 = 1}$

Left Hand Side = Right Hand Side from this Calculations. Hence Proved!