Tan(60°-A) = √3cosA -sinA÷cosA +√3sinA

1.	Tan(60°-A) = √3cosA -sinA÷cosA +√3sinA
Answer» Correct QUESTION: Prove that $\bf{tan(60^{\circ}-A)=\dfrac{\sqrt{3}cosA-sinA}{cosA+\sqrt{3}sinA}}$ . To Prove: $\longrightarrow\sf{tan(60^{\circ}-A)=\dfrac{\sqrt{3}cosA-sinA}{cosA+\sqrt{3}sinA}}$ SOLUTION: We KNOW that, $\longrightarrow\sf{tan(A-B)=\dfrac{tanA-tanB}{1+tanAtanB}}$ $\longrightarrow\sf{tan(A)=\dfrac{sinA}{cosA}}$ $\longrightarrow\sf{tan60^{\circ}=\sqrt{3}}$ Now, $\sf{L.H.S.=tan(60^{\circ}-A)}$ $\sf{L.H.S.=\dfrac{tan60^{\circ}-tanA}{1+tan60^{\circ}tanA}}$ $\sf{L.H.S.=\dfrac{\sqrt{3}-\dfrac{sinA}{cosA}}{1+\dfrac{\sqrt{3}sinA}{cosA}}}$ $\sf{L.H.S.=\dfrac{\dfrac{\sqrt{3}cosA-sinA}{\cancel{cosA}}}{\dfrac{cosA+\sqrt{3}sinA}{\cancel{cosA}}}}$ $\sf{L.H.S.=\dfrac{\sqrt{3}cosA-sinA}{cosA+\sqrt{3}sinA}}$ $\sf{L.H.S.=R.H.S.}$ Hence Proved Few more Identities: $\rightarrow\bf{sin(A+B)=sinAcosB+cosAsinB}$ $\rightarrow\bf{sin(A-B)=sinAcosB-cosAsinB}$ $\rightarrow\bf{cos(A+B)=cosAcosB-sinAsinB}$ $\rightarrow\bf{cos(A-B)=cosAcosB+sinAsinB}$ $\rightarrow\bf{tan(A+B)=\dfrac{tanA+tanB}{1-tanAtanB}}$

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