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DO THIS I WILL MARK AS BRAINLIEST ..............HURRY...................if sinθ, cosθ and tanθ are in GP then then cos^{9}θ + cos^6 θ + 3 cos^5 θ = ??? |
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Answer» Solution:- we know that if a,b,c are in gp then b^2=ac now here cos^2∅=SIN∅×tan∅ cos^2∅=sin^2∅/cos∅ cos^3∅=sin^2∅......i) now we know sin^2∅+cos^2∅=1 converting sin into cos from i) we get cos^3∅+cos^2∅=1 cubing on both SIDE we get (cos^3∅+cos^2∅)^3=1 cos^9∅+cos^6∅+3cos^3∅×cos^5∅(cos^3∅+cos^2∅)=1 putting cos^3∅=sin^2∅ from first we get cos^9∅+cos^6∅+3cos^5∅(sin^2∅+cos^2∅)=1 cos^9∅+cos^6∅+3cos^5∅-1=0 hence the value will be 0 ( my previous answer is wrong,I will CORRECT it soon) |
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