Question

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

Answer 1

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:

$\rightarrow\bf{sin(A+B)=sinAcosB+cosAsinB}$

$\rightarrow\bf{sin(A-B)=sinAcosB-cosAsinB}$

$\rightarrow\bf{cos(A+B)=cosAcosB-sinAsinB}$

$\rightarrow\bf{cos(A-B)=cosAcosB+sinAsinB}$

$\rightarrow\bf{tan(A+B)=\dfrac{tanA+tanB}{1-tanAtanB}}$