KeerthanaPosted on अगर tanθ=2\tan \theta=2tanθ=2, तो 8sinθ+5cosθsin3θ+2cos3θ+कामूल्य3cosθ\frac{8 \sin \theta+5 \cos \theta}{\sin ^{3} \theta+2 \cos ^{3} \theta+ का मूल्य 3 \cos \theta}sin3θ+2cos3θ+कामूल्य3cosθ8sinθ+5cosθ है a215\frac{21}{5}521 b85\frac{8}{5}58 c75\frac{7}{5}57 d165\frac{16}{5}516 Answer : Option AExplanation : अभिव्यक्ति =8sinθ+5cosθsin3θ+2cos2θ+3cosθ=\frac{8 \sin \theta+5 \cos \theta}{\sin ^{3} \theta+2 \cos ^{2} \theta+3 \cos \theta}=sin3θ+2cos2θ+3cosθ8sinθ+5cosθ अंश और हर को cos θ से विभाजित करना =8tanθ+5tanθ⋅sin2θ+2cos2θ+3=\frac{8 \tan \theta+5}{\tan \theta \cdot \sin ^{2} \theta+2 \cos ^{2} \theta+3}=tanθ⋅sin2θ+2cos2θ+38tanθ+5 =8tanθ+52sin2θ+2cos2θ+3=8tanθ+52(sin2θ+cos2θ)+3=\frac{8 \tan \theta+5}{2 \sin ^{2} \theta+2 \cos ^{2} \theta+3}=\frac{8 \tan \theta+5}{2\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+3}=2sin2θ+2cos2θ+38tanθ+5=2(sin2θ+cos2θ)+38tanθ+5 =8×2+55=215=\frac{8 \times 2+5}{5}=\frac{21}{5}=58×2+5=521 Rate This:NaN / 5 - 1 votesAdd comment