Answer: tanθ=2 Now, sin2θ+2cos2θ+3cosθ8sinθ+5cosθ = cos3θ×(tan3θ+2+cos2θ3)cosθ(8tanθ+5) = tan3θ+2+3sec2θsec2θ(8tanθ+5)=tan3θ+2+3(1+tan2θ)(1+tan2θ)(8tanθ+5) = 23+2+3(1+22)(1+22)(8×2+5)=10+155×21=25 Reply
Answer:
tanθ=2
Now, sin2θ+2cos2θ+3cosθ8sinθ+5cosθ
= cos3θ×(tan3θ+2+cos2θ3)cosθ(8tanθ+5)
= tan3θ+2+3sec2θsec2θ(8tanθ+5)=tan3θ+2+3(1+tan2θ)(1+tan2θ)(8tanθ+5)
= 23+2+3(1+22)(1+22)(8×2+5)=10+155×21=25