Question

The value of $\tan \dfrac{\pi }{8}\tan \dfrac{3\pi }{8}$ is
(A) $0$
(B) $1$
(C) $\dfrac{1}{2}$
(D) None of these

Answer 1

Hint: For answering this question we need to find the value of product of $\tan \dfrac{\pi }{8}$ and $\tan \dfrac{3\pi }{8}$. For finding the value of a product we need to find the value of the individual one and then the product. For finding the value of $\tan \left( \dfrac{\pi }{8} \right)$ we will use the formulae $\tan \left( \dfrac{A}{2} \right)=\sqrt{\left( \dfrac{1-\cos A}{1+\cos A} \right)}$ and for the value of $\tan \left( \dfrac{3\pi }{8} \right)$ we will use $\tan \left( \dfrac{\pi }{4}+A \right)=\dfrac{1+\tan A}{1-\tan A}$ .

Complete step by step answer:
We can write $\tan \left( \dfrac{\pi }{8} \right)$as $\tan \left( \dfrac{\pi }{4}\times \dfrac{1}{2} \right)$ for this form we can apply $\tan \left( \dfrac{A}{2} \right)=\sqrt{\left( \dfrac{1-\cos A}{1+\cos A} \right)}$ for $A={{45}^{\circ }}$ using $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ .
After applying we will have $\tan \left( \dfrac{\pi }{8} \right)=\sqrt{\left( \dfrac{1-\cos {{45}^{\circ }}}{1+\cos {{45}^{\circ }}} \right)}$ .
We can simplify this by using the value of $\cos {{45}^{\circ }}$ . After simplifying we will have $\tan \left( \dfrac{\pi }{8} \right)=\sqrt{\left( \dfrac{1-\dfrac{1}{\sqrt{2}}}{1+\dfrac{1}{\sqrt{2}}} \right)}$ .
After this we can simplify what we have and then we will have $\tan \left( \dfrac{\pi }{8} \right)=\sqrt{\left( \dfrac{\dfrac{\sqrt{2}-1}{\sqrt{2}}}{\dfrac{\sqrt{2}+1}{\sqrt{2}}} \right)}$ .
After simplifying this we will have $\tan \left( \dfrac{\pi }{8} \right)=\sqrt{\left( \dfrac{\sqrt{2}-1}{\sqrt{2}+1} \right)}$ .
For further simplifying this we will multiply and divide it by $\sqrt{2}-1$ after doing this we will have $\tan \left( \dfrac{\pi }{8} \right)=\sqrt{\left( \dfrac{\sqrt{2}-1}{\sqrt{2}+1} \right)\left( \dfrac{\sqrt{2}-1}{\sqrt{2}-1} \right)}$ .
By further simplifying this we will have $\tan \left( \dfrac{\pi }{8} \right)=\sqrt{\left( \dfrac{{{\left( \sqrt{2}-1 \right)}^{2}}}{\left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right)} \right)}$ .
And by performing further simplifications we will have $\tan \left( \dfrac{\pi }{8} \right)=\sqrt{\left( \dfrac{{{\left( \sqrt{2}-1 \right)}^{2}}}{2-1} \right)}=\sqrt{2}-1$ .
To answer this question we need to find the value of $\tan \left( \dfrac{3\pi }{8} \right)$ .
This can be written as $\tan \left( \dfrac{3\pi }{8} \right)=\tan \left( \dfrac{\pi }{4}+\dfrac{\pi }{8} \right)$ .
This can be further simplified this by using $\tan \left( \dfrac{\pi }{4}+A \right)=\dfrac{1+\tan A}{1-\tan A}$ after applying this we will have $\tan \left( \dfrac{3\pi }{8} \right)=\dfrac{1+\tan \left( \dfrac{\pi }{8} \right)}{1-\tan \left( \dfrac{\pi }{8} \right)}$ .
For simplifying this we will use $\tan \left( \dfrac{\pi }{8} \right)=\sqrt{2}-1$ after using this we will have $\tan \left( \dfrac{3\pi }{8} \right)=\dfrac{1+\sqrt{2}-1}{1-\left( \sqrt{2}-1 \right)}$ .
After simplifying this we will have $\tan \left( \dfrac{3\pi }{8} \right)=\dfrac{\sqrt{2}}{2-\left( \sqrt{2} \right)}$ .
For performing further simplifications we will take $\sqrt{2}$ as common then we will have $\tan \left( \dfrac{3\pi }{8} \right)=\dfrac{\sqrt{2}}{\left( \sqrt{2} \right)\left( \sqrt{2}-1 \right)}$ .
After this we will can simply write it as $\tan \left( \dfrac{3\pi }{8} \right)=\dfrac{1}{\left( \sqrt{2}-1 \right)}$ .
For further simplifying this we will multiply and divide it by $\sqrt{2}+1$ and after this we will have $\tan \left( \dfrac{3\pi }{8} \right)=\dfrac{1}{\left( \sqrt{2}-1 \right)}\left( \dfrac{\sqrt{2}+1}{\sqrt{2}+1} \right)=\dfrac{\sqrt{2}+1}{\left( \sqrt{2}-1 \right)\left( \sqrt{2}+1 \right)}$ .
By final conclusion we will have it as $\tan \left( \dfrac{3\pi }{8} \right)=\dfrac{\sqrt{2}+1}{2-1}=\sqrt{2}+1$ .
For answering this we will have to find the product of them $\tan \left( \dfrac{\pi }{8} \right)=\sqrt{2}-1$ and $\tan \left( \dfrac{3\pi }{8} \right)=\sqrt{2}+1$ .
We will have $\tan \left( \dfrac{\pi }{8} \right)\tan \left( \dfrac{3\pi }{8} \right)=\left( \sqrt{2}-1 \right)\left( \sqrt{2}+1 \right)$ after further simplifying we will have $\tan \left( \dfrac{\pi }{8} \right)\tan \left( \dfrac{3\pi }{8} \right)=1$ .

So, the correct answer is “Option B”.

Note: While solving this type of questions it would be efficient if we remember the value of $\tan \left( \dfrac{\pi }{8} \right)$ this is $\sqrt{2}-1$ . The value of sine and cosine values of these terms is $\sin \left( \dfrac{\pi }{8} \right)=\sqrt{\dfrac{\sqrt{2}-1}{2\sqrt{2}}}$ and $\cos \left( \dfrac{\pi }{8} \right)=\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}$ .