Math, asked by pragyanpudhari, 9 months ago

sin6A/sin2A-cos6A/cos2A

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Answered by Anonymous

34

➡(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.

Answered by Anonymous

0

ʀᴇꜰᴇʀ ᴛʜᴇ ᴀᴛᴛᴀᴄʜᴍᴇɴᴛ.....

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