Question

(cosA²)²-cos²A=(sinA²)²-sin²A

Answer 1

${\large\bf{\mid{\overline{\underline{Correct\:Question:-}}}\mid}}$

$\textbf{Prove that}$

• (cos²A)² - cos²A = (sin²A)² - sin²A

$\Large\bold\star\underline{\underline\textbf{Formula\:used}}$

• cos²A - 1 = sin²A
• sin²A - 1 = cos²A

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$\Large\underline{\underline{\sf{Solution}:}}$

$\textbf{Lets Solve LHS and RHS seperately}$

$\Large\bold\star\underline{\underline\textbf{LHS}}$

$\red \leadsto \green{(\cos^{2} (A))^{2} - \cos^{2} (A)} \\ \\ taking \: \cos^{2} (A) \: common \\ \\ \red \leadsto \: \green{\cos^{2} (A)(\cos^{2} (A) - 1)} \\ \\ taking \: ( - 1) \: common \: now \\ \\ \red\leadsto \: \green{( - 1){\cos^{2} (A)(1 - \cos^{2} (A) )}} \\ \\ \red\leadsto \: \green{( - 1){\cos^{2} (A) \: \sin^{2} (A)}} \\ \\ \red\leadsto \: \large\boxed{\bold{LHS}}$

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$\Large\bold\star\underline{\underline\textbf{RHS}}$

$\red \leadsto \pink{(\sin^{2} (A))^{2} - \sin^{2} (A)} \\ \\ taking \: \sin^{2} (A) \: common \\ \\ \red \leadsto \: \pink{\sin^{2} (A)(\sin^{2} (A) - 1)} \\ \\ taking \: ( - 1) \: common \: now \\ \\ \red\leadsto \: \pink{( - 1){\sin^{2} (A)(1 - \sin^{2} (A) )}} \\ \\ \red\leadsto \: \pink{( - 1){\sin^{2} (A) \: \cos^{2} (A)}} \\ \\ \red\leadsto \: \large\boxed{\bold{ RHS}}$

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From Both we can say that,

$\large\boxed{\bold{LHS}} = \large\boxed{\bold{ RHS}} \:$

$\LARGE\boxed{\bold{Hence,\:Proved}}$

Answer 2

Answer:

(Cos²(A)²-cos²(A)

Taking cos²(A)(cos²(A)-1)

Taking (-1) common

(-1)cos²Asin²A