Question

If A = 0° and B = 60°, find the value of tan A−tan B/1+tan Atan B. Give your answer in surd form where necessary​

Answer 1

Answer:

-√3

Step-by-step explanation:

given, A=0° and B=60°

then,

tan0° - tan60° + tan0° × tan60°

0 - √3 + 0 × √3. [tan0° = 0 and tan60°= √3]

-√3

Hope , it helps you....

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:A = 0\degree$

$\rm :\longmapsto\:B = 60\degree$

From table of Trigonometric ratios of Standard angles, we have

$\rm :\longmapsto\:tan0\degree = 0$

$\rm :\longmapsto\:tan60\degree = \sqrt{3}$

Now, Consider

$\rm :\longmapsto\:\dfrac{tanA - tanB}{1 + tanA \: tanB}$

On substituting the values of tanA and tanB, we get

$\rm \: = \: \dfrac{0 - \sqrt{3} }{1 + 0 \times \sqrt{3} }$

$\rm \: = \: \dfrac{- \sqrt{3} }{1 + 0}$

$\rm \: = \: \dfrac{- \sqrt{3} }{1}$

$\rm \: = \: - \sqrt{3}$

$\rm\implies \:\boxed{\tt{ \frac{tanA - tanB}{1 + tanA \: tanB} = - \sqrt{3} \: }}$

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