Question

If brainly is not a worse app then solve this question​

Answer 1

Step-by-step explanation:

sinA=√3/2=sin 60°

A=60°

cosB=√3/2=cos 30°

B=30°

from the above question,

=(tanA-tanB) /1+tanA.tanB

=(tan60°-tan30°)/1+tan60°×tan30°

=(√3-1/√3)/1+√3×1/√3

=2/√3×2

=1/√3 or tan30° or tanB

Answer 2

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

Given that,

$\rm :\longmapsto\:sinA = \dfrac{ \sqrt{3} }{2}$

Using Trigonometric table of some standard angles,

$\rm :\longmapsto\:sinA = sin60 \degree$

So, on Comparing we get

$\bf\implies \:A = 60 \degree$

Also, given that

$\rm :\longmapsto\:cosB = \dfrac{ \sqrt{3} }{2}$

Using Trigonometric table of some standard angles, we get

$\rm :\longmapsto\:cosB = cos30 \degree$

So, on comparing we get

$\bf\implies \:\:B = 30 \degree$

Now,

Consider,

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

We know,

$\boxed{ \sf{ \:tan(A - B) = \frac{tanA - tanB}{1 + tanAtanB}}}$

So, using this identity, we get

$\rm \: = \: \:tan(60 \degree \: - \: 30 \degree \: )$

$\rm \: = \: \:tan\: 30 \degree \:$

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

Hence,

$\: \: \: \: \: \underbrace{\boxed{ \bf{ \:\bf :\longmapsto\:\dfrac{tanA - tanB}{1 + tanA \: tanB} \: = \: \frac{1}{ \sqrt{3} } \quad}}}$

Additional Information :-

$\boxed{ \sf{ \:tan(A + B) = \frac{tanA + tanB}{1 - tanAtanB}}}$

$\boxed{ \sf{ \:sin(A + B) = sinAcosB + sinBcosA}}$

$\boxed{ \sf{ \:sin(A - B) = sinAcosB - sinBcosA}}$

$\boxed{ \sf{ \:cos(A + B) = cosAcosA - sinAsinB}}$

$\boxed{ \sf{ \:cos(A - B) = cosAcosA + sinAsinB}}$

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