Question

Question 6
The value of $(tan{1}^{\circ },tan{2}^{\circ },tan{3}^{\circ }\dots \dots tan{89}^{\circ })$ is

(A) 0
(B) 1
(C) 2
(D) $\frac{1}{2}$

Answer 1

The answer is (B).

Thinking Process
Use the transformation $tan({90}^{\circ }-\theta )=cot\theta$ from greater than trigonometric angle $tan{45}^{\circ }$ after that we use the trigonometric ratio, $cot\theta =\frac{1}{tan\theta }$

$tan{1}^{\circ },tan{2}^{\circ },tan{3}^{\circ }\dots \dots tan{89}^{\circ }$

$=tan{1}^{\circ },tan{2}^{\circ },tan{3}^{\circ }\dots \dots tan{44}^{\circ },tan{45}^{\circ },tan{46}^{\circ }\dots \dots tan{87}^{\circ },tan{88}^{\circ },tan{89}^{\circ }$

$=tan{1}^{\circ },tan{2}^{\circ },tan{3}^{\circ }\dots \dots tan{44}^{\circ },\left(1\right).tan({90}^{\circ }-{44}^{\circ })\dots \dots tan({90}^{\circ }-3^{\circ })tan({90}^{\circ }-2^{\circ }).tan({90}^{\circ }-1^{\circ })(\because tan{45}^{\circ }=1)$

$=tan{1}^{\circ },tan{2}^{\circ },tan{3}^{\circ }\dots \dots tan{44}^{\circ },\left(1\right).cot{44}^{\circ }\dots \dots cot{3}^{\circ }.cot{2}^{\circ },cot{1}^{\circ }$

$\lfloor \because tan(90-\theta )=cot\theta \rfloor$

$\because tan{45}^{\circ }=1$

$=tan{1}^{\circ },tan{2}^{\circ },tan{3}^{\circ }\dots \dots tan{44}^{\circ },\left(1\right)\dots \frac{1}{tan{44}^{\circ }}\dots \frac{1}{tan{30}^{\circ }}\dots \frac{1}{tan{1}^{\circ }}[\because cot\theta =\frac{1}{tan\theta }]$