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cos^4 A/cos^2 B + sin^4 A/sin^2 B=1 then prove that cos^4 B/cos^2 A + sin^4 A/sin^2 B=1

cos^4 A/cos^2  B   +   sin^4 A/sin^2 B=1


then prove that  cos^4 B/cos^2 A    +   sin^4 A/sin^2 B=1

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1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:Hello student, please find answer to your question
\frac{cos^{4}A}{cos^{2}B} + \frac{sin^{4}A}{sin^{2}B} = 1
\cos^{4}A.sin^{2}B} + sin^{4}A.cos^{2}B = cos^{2}B.sin^{2}B
\cos^{4}A.(1-cos^{2}B) + sin^{4}A.cos^{2}B = cos^{2}B.sin^{2}B
\cos^{4}A-cos^{4}A.cos^{2}B + sin^{4}A.cos^{2}B = cos^{2}B.sin^{2}B
\cos^{4}A+ (sin^{4}A-cos^{4}A)cos^{2}B = cos^{2}B.sin^{2}B
\cos^{4}A+ (sin^{2}A-cos^{2}A)(sin^{2}A+cos^{2}A)cos^{2}B = cos^{2}B.sin^{2}B
\cos^{4}A+ (sin^{2}A-cos^{2}A)cos^{2}B = cos^{2}B.(1-cos^{2}B)
\cos^{4}A+ (sin^{2}A-cos^{2}A)cos^{2}B = cos^{2}B-cos^{4}B
\cos^{4}B+ (sin^{2}A-cos^{2}A)cos^{2}B = cos^{2}B-cos^{4}A
\cos^{4}B+ (sin^{2}A-cos^{2}A)cos^{2}B = cos^{2}B-1+1-cos^{4}A
\cos^{4}B+ (sin^{2}A-cos^{2}A)cos^{2}B = cos^{2}B-1+(1-cos^{2}A)(1+cos^{2}A)
\cos^{4}B+ (sin^{2}A-cos^{2}A)cos^{2}B = cos^{2}B-1+(sin^{2}A)(1+cos^{2}A)\cos^{4}B+ (sin^{2}A-cos^{2}A)cos^{2}B = cos^{2}B-1+sin^{2}A+sin^{2}A.cos^{2}A\cos^{4}B+ (sin^{2}A-cos^{2}A)(1-sin^{2}B) = cos^{2}B-1+sin^{2}A+sin^{2}A.cos^{2}A\cos^{4}B+ sin^{2}A-cos^{2}A-sin^{2}A.sin^{2}B+cos^{2}A.sin^{2}B = cos^{2}B-1+sin^{2}A+sin^{2}A.cos^{2}A\cos^{4}B+1-cos^{2}A-sin^{2}A.sin^{2}B+cos^{2}A.sin^{2}B = cos^{2}B+sin^{2}A.cos^{2}A
\cos^{4}B+1-cos^{2}B-cos^{2}A-sin^{2}A.sin^{2}B+cos^{2}A.sin^{2}B = sin^{2}A.cos^{2}A\cos^{4}B+sin^{2}B-cos^{2}A-sin^{2}A.sin^{2}B+cos^{2}A.sin^{2}B = sin^{2}A.cos^{2}A\cos^{4}B+sin^{2}B(1-sin^{2}A)-cos^{2}A+cos^{2}A.sin^{2}B = sin^{2}A.cos^{2}A\cos^{4}B+sin^{2}B.cos^{2}A-cos^{2}A+cos^{2}A.sin^{2}B = sin^{2}A.cos^{2}A
cos^{4}B-cos^{2}B.cos^{2}A+cos^{2}A.sin^{2}B = sin^{2}A.cos^{2}A
cos^{4}B+cos^{2}A(sin^{2}B-cos^{2}B) = sin^{2}A.cos^{2}A
\Rightarrow \frac{cos^{4}B}{cos^{2}A} + \frac{sin^{4}B}{sin^{2}A} = 1

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