Question

prove that tan²A-sin²A=tan²A•sin²A

Answer 1

= (sin²A | cos²A) - sin²A 
= (sin²A - cos²A • sin²A) | cos²A 
= sin²A [1-cos²A] | cos²A 
= Sin²A [sin²A] | cos²A 
= Sin²A [sin²A | cos²A] 
= sin²A • tan²A = RHS

Answer 2

$\huge \underline \mathfrak\red{Question}$

prove that tan²A-sin²A=tan²A•sin²A

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$\large \underline \green{ \rm \: L.H.S}= \tan²(A) -\sin²(A)$

$\large \underline \green{ \rm \: R.H.S}= \tan²(A) × \sin²(A)$

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$\large \underline{ \rm \: Now \: let's \: prove}$

$\large \underline \green{ \rm \: L.H.S}:-$

$\dfrac{ \ \sin ²(A) }{ \cos²(A) } = \tan²(A)$

$→\dfrac{ \sin²(A) }{ \cos²(A) } - \sin²(A)$

$→ \dfrac{ \sin²(A)- \cos²(A)× \sin²(A) }{ \cos²(A) }$

$→ \dfrac{ \sin²A[1- \cos²(A)] }{ \cos²(A) }$

$\boxed { \orange{1 - \cos²(A) = \sin²(A) }}$

$→ \dfrac{ \sin²(A)[ \sin²(A)] }{ \cos²(A) }$

$→ \sin²(A) \: \: [ \frac{ \sin²(A) }{ \cos²(A) } ]$

$\boxed { \orange{→ \frac{ \sin²(A) }{ \cos²(A) } = \tan²(A) }}$

$\sin²(A) × \tan²(A) =\large \underline \green{ \rm \: R.H.S}$

$\large \underline \green{ \rm \: L.H.S=R.H.S}$

$\large \underline \green{ \rm \: Hence \: proved}$

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$\rm \blue{@BrainlyBlockBusterBB}$