Question

if sin A=√3/2 and cos B=√3/2, find the value of: tanA-tanB/1+tan A tan B

Answer 1

SinA=Sin60 => A=60
CosB=Cos30 =>B=30
Now the given expression can be compactly written as
Tan(A-B) as it is a formula.
Therefore Tan(60-30)=Tan30=1/root3

Answer 2

The value of $\dfrac{\tan A-\tan B}{1+\tan A \tan B}=\dfrac{1}{\sqrt{3}}$.

Step-by-step explanation:

We have,

$\sin A=\dfrac{\sqrt{3}}{2}$ and $\cos B=\dfrac{\sqrt{3}}{2}$

To find, the value of $\dfrac{\tan A-\tan B}{1+\tan A \tan B}$=?

∴ $\sin A=\dfrac{\sqrt{3}}{2}$

⇒ $\sin A=\sin 60$ [ ∵ $\sin 60=\dfrac{\sqrt{3}}{2}$]

⇒ A = 60°

Also,

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

⇒ $\cos B=\cos 30$ [ ∵ $\cos 30=\dfrac{\sqrt{3}}{2}$]

⇒ B = 30°

∴ $\dfrac{\tan A-\tan B}{1+\tan A \tan B}$

$=\dfrac{\tan 60-\tan 30}{1+\tan 60 \tan 30}$

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

[ ∵ $\tan 60=\sqrt{3}$ and $\tan 30=\dfrac{1}{3}$]

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

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

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

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

Hence, the value of $\dfrac{\tan A-\tan B}{1+\tan A \tan B}=\dfrac{1}{\sqrt{3}}$