sin 52°+cot38°\sin 38°+cos 52°=tan 52°
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Answered by
2
sin52+cot(90-38)/sin38+cos(90-52)=tan52
=sin52+sin52/sin38+sin38=tan52
=0
=sin52+sin52/sin38+sin38=tan52
=0
Answered by
5
Sin52°+Cos38°/Sin38°+Cis52° = Tan52°
LHS = SIN 52°+COS 38°/SIN38°+COS 52°
Sin52°+ Cos(90-52°)/Sin38°+Cos(90-38)°
Sin52° + Sin52° / Sin(90-52°)+ Cos52° [ Cos(90-theta=Sin theta]>
2Sin52°/2Cos52° = Sin52°/Cos 52°
SIN52°/COS 52° = TAN52° [ SIN THETA/ COS THETA= TAN THETA]
HENCE,
LHS = RHS
LHS = SIN 52°+COS 38°/SIN38°+COS 52°
Sin52°+ Cos(90-52°)/Sin38°+Cos(90-38)°
Sin52° + Sin52° / Sin(90-52°)+ Cos52° [ Cos(90-theta=Sin theta]>
2Sin52°/2Cos52° = Sin52°/Cos 52°
SIN52°/COS 52° = TAN52° [ SIN THETA/ COS THETA= TAN THETA]
HENCE,
LHS = RHS
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