If \frac{\cos^4 \alpha}{\cos ^2 \beta}+\frac{\sin^4 \alpha}{\sin^2 \beta}=1 then the value of \frac{\cos^4 \beta}{\cos ^2 \alpha}+\frac{\sin^4 \beta}{\sin^2 \alpha} is

RRB Junior Engineer Stage 1 Model Papers 6

Question : 17 of 99

Marks: +1, -0

{\cos^4 α}/{\cos ^2 β}+{\sin^4 α}/{\sin^2 β}=1

then the value of

{\cos^4 β}/{\cos ^2 α}+{\sin^4 β}/{\sin^2 α}

Solution:

{\cos^4 α}/{\cos ^2 β}+{\sin^4 α}/{\sin^2 β}=1

⇒{\cos^4 α \;\sin^2 β+\sin^4 α\;\cos ^2 β=\cos ^2 β\;\sin^2 β

⇒\cos^4 α(1-\cos ^2 β)+\cos ^2 β(1-\cos ^2 α)^2=\cos ^2 β(1-\cos ^2 β)

⇒\cos^4 α-\cos^4 α\;\cos^2 β+\cos^2 β-2\cos^2 α \;\cos^2 β+\cos^4 α\;\cos^2 β=\cos^2 β-\cos^4 β

⇒\cos^4 α-2\cos^2 α\;\cos^2β+\cos^4 β=0

⇒(\cos^2 α-\cos^2 β)^2=0

⇒\cos^2 α=\cos^2 β

⇒\sin^2 α=\sin^2 β

Then,

{\cos^4 β}/{\cos ^2 α}+{\sin^4 β}/{\sin^2 α}={\cos ^2 β\;\cos ^2 α}/{\cos ^2 α}+{\sin^2 β\;\sin^2 α}/{\sin^2 α}

⇒\cos^2β+\sin^2 β=1

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