Question

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

• A. $-1$
• B. $1$
• C. $-3$
• D. $3$

Answer 1

Answer: (C), (D)

$-3$
$3$

Given $\tan 1^o\tan 2^o\tan 3^o..........\tan {87}^o\tan {88}^o\tan {89}^o=x^2-8$

$\tan 1^o\tan 2^o\tan 3^o..........\tan {87}^o\tan {88}^o\tan {89}^o$

$=\left(\tan 1^o\tan {89}^o\right)\left(\tan 2^o\tan {88}^o\right)...\left(\tan {44}^o\tan {46}^o\right)\tan {45}^o$
$=\left(\tan 1^o\tan \right(90-1{)}^o\left)\right(\tan 2^o\tan (90-2{)}^o).......\left(\tan {44}^o\tan \right(90-44{)}^o)\tan {45}^o$

we know that $\cot A=\tan (90˚-A)$

$=\left(\tan 1^o\cot 1^o\right)\left(\tan 2^o\cot 2^o\right).......\left(\tan {44}^o\cot {44}^o\right)\tan {45}^o=1$

Hence, $1=x^2-8\Rightarrow x^2=9\Rightarrow x=\pm 3$

Therefore, Answer is $-3$ and $3$