Question

Find the value of the following product:$\tan {{10}^{o}}\tan {{20}^{o}}\tan {{30}^{o}}\tan {{40}^{o}}\tan {{50}^{o}}\tan {{60}^{o}}\tan {{70}^{o}}\tan {{80}^{o}}$
A. $0$
B. $-1$
C. $\dfrac{1}{\sqrt{3}}$
D. $1$

Answer 1

The given problem is related to trigonometry. Try to remember the basic values of sine, cosine, and tangent of angles which are of the form ${{90}^{o}}-\theta$. and then by solving you can find the answer.
Complete step-by-step Solution:
Let us consider two angles $\alpha$ and $\beta$ such that $\alpha +\beta ={{90}^{o}}$. Here, $\alpha$ and $\beta$ are called complementary angles.
Now, we know, $\sin \left( {{90}^{o}}-\theta \right)=\cos \theta$ and $\cos \left( {{90}^{o}}-\theta \right)=\sin \theta$.
Now, when we consider the angles $\alpha$ and $\beta$ such that $\alpha +\beta ={{90}^{o}}$ , we get $\alpha ={{90}^{o}}-\beta$.
So, $\sin \alpha =\sin \left( {{90}^{o}}-\beta \right)=\cos \beta ......(i)$ and $\cos \alpha =\cos \left( {{90}^{o}}-\beta \right)=\sin \beta .....(ii)$.
Now, we know, $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$.
So, $\cot \alpha =\dfrac{\cos \alpha }{\sin \alpha }....(iii)$.
Now, we will substitute equations $(i)$ and $(ii)$in equation$(iii)$.
On substituting equations $(i)$ and $(ii)$in equation$(iii)$, we get
$\cot \alpha =\dfrac{\cos \left( {{90}^{o}}-\beta \right)}{\sin \left( {{90}^{o}}-\beta \right)}$
So, $\cot \alpha =\dfrac{\sin \beta }{\cos \beta }$
Or, $\cot \alpha =\dfrac{1}{\cot \beta }.....(iv)$
Now, we know $\cot \beta =\dfrac{1}{\tan \beta }$.
We will take $\tan \beta$ to the left-hand side of the equation and $\cot \beta$ to the right-hand side of the equation.
So, we get $\tan \beta =\dfrac{1}{\cot \beta }.....(v)$.
Now, we will substitute equation $(v)$ in equation$(iv)$.
On substituting equation $(v)$ in equation$(iv)$, we get $\cot \alpha =\tan \beta ...(vi)$.
Now, we will consider the value of $\alpha$ to be equal to ${{10}^{o}}$.
Now, we know, $\beta ={{90}^{o}}-\alpha$ .
So, when the value of $\alpha$ is equal to ${{10}^{o}}$, then the value of $\beta$ is given as $\beta ={{90}^{o}}-{{10}^{o}}={{80}^{o}}$.
Now, from equation$(vi)$, we have $\cot \alpha =\tan \beta$.
We will substitute the values of $\alpha$ and $\beta$ in equation$(vi)$.
On substituting the values of $\alpha$ and $\beta$ in equation$(vi)$, we get $\tan {{80}^{o}}=\cot {{10}^{o}}....(vii)$.
Now, we will consider the value of $\alpha$ to be equal to ${{20}^{o}}$.
Now, we know, $\beta ={{90}^{o}}-\alpha$ .
So, when the value of $\alpha$ is equal to ${{20}^{o}}$, then the value of $\beta$ is given as $\beta ={{90}^{o}}-{{20}^{o}}={{70}^{o}}$.
Now, from equation$(vi)$, we have $\cot \alpha =\tan \beta$.
We will substitute the values of $\alpha$ and $\beta$ in equation$(vi)$.
On substituting the values of $\alpha$ and $\beta$ in equation$(vi)$, we get $\tan {{70}^{o}}=\cot {{20}^{o}}....(viii)$.
Similarly, $\tan {{60}^{o}}=\tan \left( {{90}^{o}}-{{30}^{o}} \right)=\cot {{30}^{o}}.....(ix)$ , and $\tan {{50}^{o}}=\tan \left( {{90}^{o}}-{{40}^{o}} \right)=\cot {{40}^{o}}......(x)$ .
Now, we are asked to find the value of $\tan {{10}^{o}}\tan {{20}^{o}}\tan {{30}^{o}}\tan {{40}^{o}}\tan {{50}^{o}}\tan {{60}^{o}}\tan {{70}^{o}}\tan {{80}^{o}}$. From equation $\left( vii \right),\left( viii \right),\left( ix \right)$ and $\left( x \right)$ , we get:
$\tan {{10}^{o}}\tan {{20}^{o}}\tan {{30}^{o}}\tan {{40}^{o}}\tan {{50}^{o}}\tan {{60}^{o}}\tan {{70}^{o}}\tan {{80}^{o}}$
$=\tan {{10}^{o}}\tan {{20}^{o}}\tan {{30}^{o}}\tan {{40}^{o}}\cot {{40}^{o}}\cot {{30}^{o}}\cot {{20}^{o}}\cot {{10}^{o}}$
$=\tan {{10}^{o}}\cot {{10}^{o}}\tan {{20}^{o}}\cot {{20}^{o}}\tan {{30}^{o}}\cot {{30}^{o}}\tan {{40}^{o}}\cot {{40}^{o}}$
$=1\times 1\times 1\times 1$
$=1$
So, the value of $\tan {{10}^{o}}\tan {{20}^{o}}\tan {{30}^{o}}\tan {{40}^{o}}\tan {{50}^{o}}\tan {{60}^{o}}\tan {{70}^{o}}\tan {{80}^{o}}$ is $1$ .
Hence, option D. is the correct answer.

Note: Students generally get confused between $\cos \left( {{90}^{o}}-\theta \right)=\sin \theta$and $\cos \left( {{90}^{o}}+\theta \right)=-\sin \theta$. Sometimes, by mistake students write $\cos \left( {{90}^{o}}-\theta \right)$ as $-\sin \theta$ and $\cos \left( {{90}^{o}}+\theta \right)$ as $\sin \theta$, which is wrong . Sign mistakes are common but can result in getting a wrong answer. Hence, students should be careful while using trigonometric formulae and should take care of sign conventions.