Question

The value of the expression $(\tan 1^0\tan 2^0\tan 3^0...\tan {89}^0)$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$1$

Given:

$A=\tan 1^o\cdot \tan 2^o\cdot \tan 3^o\cdots \cdots \tan {89}^o$

We know that, the value of $\tan {89}^o=\tan ({90}^o-1^o)=\cot 1^o$
In the same way,
$\tan {88}^o=\cot 2^o$
$\tan {87}^o=\cot 3^o$
And .... so on, up to $\tan {46}^o=\cot {44}^o$
And the middle one is $\tan {45}^o=1$
So,

$A=\tan 1^o\cdot \tan 2^o\cdot \tan 3^o\cdots \cdots \tan {44}^o\cdot \tan {45}^o\cdot \cot {44}^o\cdots \cdots \cot 2^o\cdot \cdot \cot 1^o$
$\tan$ and $\cot$ cancels out each other, then $\tan {45}^o=1$ is remaining

$\therefore$ The value of the expression $V$ is $1$.

Hence, option B.