Question

If $\tan 1^o\tan 2^o...\tan {89}^o=x^2-8$, then the value of $x$ can be

• A. $-1$
• B. $1$
• C. $-3$
• D. $3$

Answer 1

Answer: (C), (D)

The correct options are
C $-3$
D $3$

Given that,

$\tan 1^o\tan 2^o...\tan {89}^o=x^2-8$

To find out,

The value of $x$

We can rewrite the given equation as:

$x^2-8=\left(\tan 1^o\tan {89}^o\right)\left(\tan 2^o\tan {88}^o\right)....\left(\tan {44}^o\tan {46}^o\right)\tan {45}^o$

We can write $\tan {89}^o$ as $\tan ({90}^o-1^o),\tan {88}^o$ as $\tan ({90}^o-2^o)$ and so on up to $\tan {46}^o$ as $\tan ({90}^o-{44}^o)$

Hence, the equation becomes:

$x^2-8=\left(\tan 1^o\tan \right({90}^o-1^o\left)\right)\left(\tan 2^o\tan \right({90}^o-2^o\left)\right)....\left(\tan {44}^o\tan \right({90}^o-{44}^o\left)\right)\tan {45}^o$

We know that, $\tan ({90}^o-\theta )=\cot \theta$

So, $x^2-8=\left(\tan 1^o\cot 1^o\right)\left(\tan 2^o\cot 2^o\right)....\left(\tan {44}^o\cot {44}^o\right)\tan {45}^o$

Also, $\cot \theta =\frac{1}{\tan \theta }$

So, $x^2-8=(\tan 1^o\times \frac{1}{\tan 1^o})(\tan 2^o\times \frac{1}{\tan 2^o})....(\tan {44}^o\times \frac{1}{\tan {44}^o})\tan {45}^o$

$x^2-8=\left(1\right)\left(1\right)....\left(1\right)\times \tan {45}^o$

$\tan {45}^o=1$

So, $x^2-8=\left(1\right)\left(1\right)....\left(1\right)\times 1$

$\Rightarrow x^2-8=1$

$\Rightarrow x^2=9$

$\Rightarrow x=\pm 3$

Hence, if $\tan 1^o\tan 2^o...\tan {89}^o=x^2-8$, the value of $x$ is $3$ or $-3$.