If 5[tan^2x-cos^2x]=2cos2x+9 then find the value of cos4x a).1/3 b). 2/9 c).-7/9 d).-3/5

Question

Ayanava Datta · Answer

5(tan 2 x-cos 2 x)=2cos2x+9 (5/a 2) -5a 2 -14=2(2a 2 -1) [a=cosx] 5-9a 4 -12a 2 =0 9a 4 +12a 2 -5=0 (3a 2 +5)(3a 2 -1)=0 a 2 ​=(-5/3)/(1/3) a squre can’t be -ve a 2 =(1/3) cos 2 x=(1/3) cos4x=2(2cos 2 x-1) 2 -1 =(-7/9) ans: (c)

ankit singh · Answer

If 5 ( tan ^ 2 x – cos^ 2x ) = 2 cos 2x + 9 , then the value of cos 4x is ... - 3/5 ; 1/3 ; 2/9 ; - 7/ 9 ... 5 (sec 2 x – 1 – cos 2 x ) = 2(2cos 2 x – 1) + 9 ... - 7/9 . thank u regards ankit singh manit bhopal

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If 5[tan^2x-cos^2x]=2cos2x+9 then find the value of cos4x a).1/3 b). 2/9 c).-7/9 d).-3/5

If 5[tan^2x-cos^2x]=2cos2x+9 then find the value of cos4x
a).1/3 b). 2/9 c).-7/9 d).-3/5

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If 5[tan^2x-cos^2x]=2cos2x+9 then find the value of cos4x a).1/3 b). 2/9 c).-7/9 d).-3/5

If 5[tan^2x-cos^2x]=2cos2x+9 then find the value of cos4xa).1/3 b). 2/9 c).-7/9 d).-3/5

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If 5[tan^2x-cos^2x]=2cos2x+9 then find the value of cos4x
a).1/3 b). 2/9 c).-7/9 d).-3/5