Question

sin6A/sin2A-cos6A/cos2A​

Answer 1

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➡(Sin6α+sin2α) / (cos6α+cos2α) =( 2sin(6α+2α)/2* cos(6α- 2α)/2 ) /

➡( 2cos(6α+2α)/2* cos(6α- 2α)/2 ) =sin4α*cos2α / cos4α*cos2α =tq4α.

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Answer 2

Step-by-step explanation:

(Sin6α+sin2α) / (cos6α+cos2α) =( 2sin(6α+2α)/2* cos(6α- 2α)/2 ) /

➡( 2cos(6α+2α)/2* cos(6α- 2α)/2 ) =sin4α*cos2α / cos4α*cos2α =tq4α