Question

How do you use the sum or difference identities to find the exact value of $ \tan \left( \dfrac{23\pi }{12} \right) $ ?

Answer 1

Hint: We know that $ \tan \left( 2\pi -\theta \right)=-\tan \theta $ We can use this formula to solve the above question.
We can see the value of $ \dfrac{23\pi }{12}=2\pi -\dfrac{\pi }{12} $ so if we can calculate the value of $ \tan \left( \dfrac{\pi }{12} \right) $ then we easily can calculate the value of $ \tan \left( \dfrac{23\pi }{12} \right) $ .

Complete step by step answer:
We have find the value of $ \tan \left( \dfrac{23\pi }{12} \right) $ . we know the property of tan x that $ \tan \left( 2\pi -x \right)=-\tan \left( x \right) $
We can write
 $ \tan \left( \dfrac{23\pi }{12} \right)=\tan \left( 2\pi -\dfrac{\pi }{12} \right) $
 $ \Rightarrow \tan \left( \dfrac{23\pi }{12} \right)=-\tan \dfrac{\pi }{12} $
So now we have to calculate the value of $ \tan \dfrac{\pi }{12} $
We calculate the value by difference identities
We can write $ \tan \dfrac{\pi }{12}=\tan \left( \dfrac{\pi }{4}-\dfrac{\pi }{6} \right) $
We know the formula for $ \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} $
By applying this formula to above equation we get
 $ \Rightarrow \tan \dfrac{\pi }{12}=\dfrac{\tan \dfrac{\pi }{4}-\tan \dfrac{\pi }{6}}{1+\tan \dfrac{\pi }{4}\tan \dfrac{\pi }{6}} $
We know that the value of $ \tan \dfrac{\pi }{4} $ is 1 and the value of $ \tan \dfrac{\pi }{6} $ is $ \dfrac{1}{\sqrt{3}} $ by replacing these values in the above equation we get
 $ \Rightarrow \tan \dfrac{\pi }{12}=\dfrac{1-\dfrac{1}{\sqrt{3}}}{1+1.\dfrac{1}{\sqrt{3}}} $
By further solving we get
   $ \Rightarrow \tan \dfrac{\pi }{12}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1} $
 By multiplying $ \sqrt{3}+1 $ in both denominator and numerator we get
So $ \tan \dfrac{\pi }{12}=\dfrac{4-2\sqrt{3}}{2} $
That gives us $ \tan \dfrac{\pi }{12}=2-\sqrt{3} $
 $ \tan \left( \dfrac{23\pi }{12} \right)=-\tan \dfrac{\pi }{12} $
Now we can see that the value of $ \tan \left( \dfrac{23\pi }{12} \right) $ is equal to $ \sqrt{3}-2 $.

Note:
We also can calculate the value of $ \tan \left( \dfrac{23\pi }{12} \right) $ by using sum identities we know the property of tan x that $ \tan \left( \pi +x \right)=\tan x $ . so we can write $ \tan \left( \dfrac{23\pi }{12} \right)=\tan \left( \pi +\dfrac{11\pi }{12} \right)=\tan \left( \dfrac{11\pi }{12} \right) $
Now we have to calculate the value of $ \tan \left( \dfrac{11\pi }{12} \right) $
We can write $ \tan \left( \dfrac{11\pi }{12} \right)=\tan \left( \dfrac{2\pi }{3}+\dfrac{\pi }{4} \right) $
We know the property
 $ \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} $