The geometric mean of tan 1^{circ}, tan 2^{ci | vedclass

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

Question

(B) Let $P = 	an 1^{\circ} \cdot 	an 2^{\circ} \cdot \ldots \cdot 	an 89^{\circ}$.Using the property $	an(90^{\circ} - 	heta) = \cot 	heta$,we c

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let $P = 	an 1^{\circ} \cdot 	an 2^{\circ} \cdot \ldots \cdot 	an 89^{\circ}$.Using the property $	an(90^{\circ} - 	heta) = \cot 	heta$,we can write the product as:$P = (	an 1^{\circ} \cdot 	an 89^{\circ}) \cdot (	an 2^{\circ} \cdot 	an 88^{\circ}) \cdot \ldots \cdot (	an 44^{\circ} \cdot 	an 46^{\circ}) \cdot 	an 45^{\circ}$.Since $	an 	heta \cdot 	an(90^{\circ} - 	heta) = 	an 	heta \cdot \cot 	heta = 1$,each pair equals $1$.Thus,$P = 1 \cdot 1 \cdot \ldots \cdot 1 \cdot 	an 45^{\circ} = 1 \cdot 1 = 1$.The geometric mean of $n$ terms $a_1, a_2, \ldots, a_n$ is $(a_1 \cdot a_2 \cdot \ldots \cdot a_n)^{1/n}$.Here,$n = 89$,so the geometric mean is $(P)^{1/89} = (1)^{1/89} = 1$.Hence,option $B$ is correct.

The geometric mean of $\tan 1^{\circ}, \tan 2^{\circ}, \ldots, \tan 89^{\circ}$ is

Similar Questions

The graph of the function $f(x) = \cos x \cos(x + 2) - \cos^2(x + 1)$ is:

If $\tan B = \frac{n \sin A \cos A}{1 - n \cos^2 A}$,then $\tan(A + B)$ equals

If $0 < x < \pi$ and $\cos x + \sin x = \frac{1}{2}$,then $\tan x$ is:

$16 \sin 12^{\circ} \cos 18^{\circ} \sin 48^{\circ} = $

If $\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$,then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module