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 $x = \tan 15^{\circ}$,$y = \operatorname{cosec} 75^{\circ}$ and $z = 4 \sin 18^{\circ}$,then :

Prove that $\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$.

Find the values of the other five trigonometric functions if $\sec x = \frac{13}{5}$ and $x$ lies in the fourth quadrant.

$1+\sec ^2 x \sin ^2 x=$

$\frac{\sin 70^\circ + \cos 40^\circ}{\cos 70^\circ + \sin 40^\circ} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module