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find the exact value of tan40deg. - tan10deg./1+tan40deg.*tan10deg.
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this is in the form of tan(x-y)... \[\tan(40-10)= \frac{1}{\sqrt3}\]
it is a common trig identity
yep
agree
is this the same as squarert3/3 ??
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\[\frac{\tan (40 {}^{\circ})-\tan (10 {}^{\circ})}{\tan (10 {}^{\circ}) \tan (40 {}^{\circ})+1}=0.57735 \]
\[\frac{1}{\sqrt3}=\frac{\sqrt{3}}{3}\]
thank you. I asked that question after you gave me it, I figured that was correct. Thank you!
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