sumlimits_{r = 1}^{89} {{log _3}(tan {r^circ} | vedclass

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

Question

(D) Let $S = \sum\limits_{r = 1}^{89} {{\log _3}(	an {r^\circ})}$.Using the property $\log_a x + \log_a y = \log_a (xy)$,we get $S = \log_3 (	an 1^\

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let $S = \sum\limits_{r = 1}^{89} {{\log _3}(	an {r^\circ})}$.Using the property $\log_a x + \log_a y = \log_a (xy)$,we get $S = \log_3 (	an 1^\circ \cdot 	an 2^\circ \cdot \dots \cdot 	an 89^\circ)$.We know that $	an(90^\circ - 	heta) = \cot 	heta$,so $	an 89^\circ = \cot 1^\circ$,$	an 88^\circ = \cot 2^\circ$,and so on.The product $P = 	an 1^\circ \cdot 	an 2^\circ \cdot \dots \cdot 	an 44^\circ \cdot 	an 45^\circ \cdot 	an 46^\circ \cdot \dots \cdot 	an 89^\circ$.Since $	an 	heta \cdot \cot 	heta = 1$,the product of terms $	an r^\circ \cdot 	an(90^\circ - r^\circ) = 1$ for $r = 1$ to $44$.Thus,$P = (1 \cdot 1 \cdot \dots \cdot 1) \cdot 	an 45^\circ = 1 \cdot 1 = 1$.Therefore,$S = \log_3 (1) = 0$.

$\sum\limits_{r = 1}^{89} {{\log _3}(\tan {r^\circ})} = $

Similar Questions

If $\theta$ is in the interval $\left(0, \frac{\pi}{2}\right)$ satisfying the equation $\cos 2 \theta \cdot \sec ^4 \theta + \sec ^2 \theta = 0$,then $\sin ^2 \theta =$

If $\cosh 2x = 199$,then $\coth x$ equals

$\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5}$ is equal to

Let $S = \{x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) : 9^{1-\tan^2 x} + 9^{\tan^2 x} = 10\}$ and $\beta = \sum_{x \in S} \tan^2\left(\frac{x}{3}\right)$,then $\frac{1}{6}(\beta - 14)^2$ is equal to

If $x = a \cos^3 \theta$ and $y = b \sin^3 \theta$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module