Question

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 = ???

Answer 1

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)

Answer 2

HERE'S THE ANSWER IN THE ATTACHMENT

hope you'll get it ✌️✌️✌️

TC :-)