If sin4x/2 + cos​4x /3 = 1/5 then
(a) tan2x = 2/3
(b) sin8x / 8 + cos8x / 27 = 1 / 125
(c) tan2x =1/3
(d)​sin8x / 8 + cos8x / 27 = 2 / 125

Dear Student,
Please find below the solution to the asked query:

Given thatsin4x2+cos4x3=15sin4x2+1-sin2x23=15Let sinx=tt42+1-t223=153t4+21-t226=1515t4+101-t22=615t4+101+t4-2t2-6=015t4+10+10t4-20t2-6=025t4-20t2+4=05t22-2.2.5t2+22=05t2-22=05t2-2=0t2=25sin2x=251-cos2x=25cos2x=1-25cos2x=35tan2x=sin2xcos2x=2535=23tan2x=23Answersin2x=25sin2x4=254sin8x=2×8625sin8x8=2625cos2x=35cos2x4=354cos8x=3×27625cos8x27=3625sin8x8+cos8x27=2625+3625=5625sin8x8+cos8x27=1125AnswerHence optiona,b are correct.

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