tan 1^{circ} times tan 2^{circ} times tan 3^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know that $	an(90^{\circ} - 	heta) = \cot 	heta$. The given expression is $P = 	an 1^{\circ} 	an 2^{\circ} 	an 3^{\circ} \cdots 	an 44^{

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) We know that $	an(90^{\circ} - 	heta) = \cot 	heta$. The given expression is $P = 	an 1^{\circ} 	an 2^{\circ} 	an 3^{\circ} \cdots 	an 44^{\circ} 	an 45^{\circ} 	an 46^{\circ} \cdots 	an 88^{\circ} 	an 89^{\circ}$. We can write $	an 89^{\circ} = 	an(90^{\circ} - 1^{\circ}) = \cot 1^{\circ}$,$	an 88^{\circ} = \cot 2^{\circ}$,and so on. Thus,$P = (	an 1^{\circ} \cot 1^{\circ}) 	imes (	an 2^{\circ} \cot 2^{\circ}) 	imes \cdots 	imes (	an 44^{\circ} \cot 44^{\circ}) 	imes 	an 45^{\circ}$. Since $	an 	heta 	imes \cot 	heta = 1$ and $	an 45^{\circ} = 1$,we have $P = 1 	imes 1 	imes \cdots 	imes 1 	imes 1 = 1$.

$\tan 1^{\circ} \times \tan 2^{\circ} \times \tan 3^{\circ} \times \cdots \times \tan 89^{\circ} = $

Similar Questions

If $\sin \left( x + \frac{4\pi}{9} \right) = a$ and $\frac{\pi}{9} < x < \frac{\pi}{3}$,then $\cos \left( x + \frac{7\pi}{9} \right)$ equals:

If $\cot \theta = -\frac{2}{3}$ and $\theta$ does not lie in the $4^{\text{th}}$ quadrant,then $\frac{(5 \sin \theta + \cos \theta)^2}{\tan \theta + \cot \theta} = $

If $\sin (\alpha - \beta ) = \frac{1}{2}$ and $\cos (\alpha + \beta ) = \frac{1}{2},$ where $\alpha$ and $\beta$ are positive acute angles,then:

The degree measure of $\frac{\pi}{32}$ radians is equal to:

$\frac{1}{\cos 290^{\circ}}+\frac{1}{\sqrt{3} \sin 250^{\circ}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module