please solve it.By using trigonometry formulas
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★ TRIGONOMETRIC REDUCTION ★
√3COT40° - 4 COS40°
√3COS40°/SIN 40° - 4COS40°
2[ √3/2COS 40°-SIN 80°] /SIN 40°
[2SIN 60° COS 40° - 2SIN80°]/ SIN 40°
SIN 100° + SIN 20° - 2SIN80° / SIN 40°
[SIN 20°-SIN 80°]/SIN 40°
-2COS 50°SIN 30° /SIN 40°
COS 50°=SIN 40°
THE GIVEN EQUATION IS REDUCED TO
-2 [ SIN 30° ]
-1
★✩★✩★✩★✩★✩★✩★✩★✩★✩★★✩★✩★
√3COT40° - 4 COS40°
√3COS40°/SIN 40° - 4COS40°
2[ √3/2COS 40°-SIN 80°] /SIN 40°
[2SIN 60° COS 40° - 2SIN80°]/ SIN 40°
SIN 100° + SIN 20° - 2SIN80° / SIN 40°
[SIN 20°-SIN 80°]/SIN 40°
-2COS 50°SIN 30° /SIN 40°
COS 50°=SIN 40°
THE GIVEN EQUATION IS REDUCED TO
-2 [ SIN 30° ]
-1
★✩★✩★✩★✩★✩★✩★✩★✩★✩★★✩★✩★
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