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

Hint: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.