Question

if A=60^o,B=30^o then prove that tan(A-B) =tanA-tanB/1+tanA.tanB​

Answer 1

Refer The Attachment ⬆️

Hence Proved,

$\boxed{\blue{\sf{Tan(A-B) = \frac{TanA-TanB}{1+TanA.TanB}}}}$

$\Large{✓}$

Hope It Helps You ✌️

Answer 2

Given :-

• A = 60°
• B = 30°

To prove :-

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

SOLUTION:-

Take L.H.S

${tan(A-B)}$

${tan(60^{\circ} - 30^{\circ})}$

${tan(30^{\circ})}$

$\dfrac{ 1}{ \sqrt{3} }$

$L.H.S = \dfrac{ 1 }{ \sqrt{3} }$

Now , take R.H.S

$\dfrac{tanA-tanB}{1+tanA tanB}$

$\dfrac{tan60^{\circ} - tan30^{\circ}}{1+tan60^{\circ}tan30^{\circ}}$

As all we know that

${tan60^{\circ}} = \sqrt{3}$

${tan30^{\circ}} = \dfrac{1}{\sqrt{3}}$

$= \dfrac{ \sqrt{3} - \dfrac{ 1 }{ \sqrt{3} } }{1 + \sqrt{3} \times \dfrac{ 1 }{ \sqrt{3} } }$

$= \dfrac{ \dfrac{ \sqrt{3} \times \sqrt{3} - 1 }{ \sqrt{3} } }{1 + 1}$

$= \dfrac{ \dfrac{3 - 1}{ \sqrt{3} } }{2}$

$= \dfrac{ \dfrac{2}{ \sqrt{3} } }{2}$

$= \dfrac{2}{2 \sqrt{3} }$

$= \dfrac{1}{ \sqrt{3} }$

$R.H.S= \dfrac{1}{ \sqrt{3} }$

Since ,

$\dfrac{1}{\sqrt{3}} = \dfrac{1}{\sqrt{3}}$

L.H.S = R.H.S

Verified!

So,

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

Know more:-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$