Math Question and Answers


Question and Answers Forum

All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 164703 by amin96 last updated on 20/Jan/22

	Answered by mr W last updated on 20/Jan/22
	$Σ_(n=1) ^(997) (tan^2 x_n +(1/(tan^2 x_n ))) =Σ_(n=1) ^(997) (tan^2 x_n +(1/(tan^2 x_n ))−2+2) =Σ_(n=1) ^(997) {(tan x_n −(1/(tan x_n )))^2 +2} ≥Σ_(n=1) ^(997) {2}=997×2=1994 = is valid only if tan x_n −(1/(tan x_n ))=0, i.e. tan x_n =(1/(tan x_n )) ⇒tan x_n =±1 ⇒x_n =k_n π±(π/4)$
	$\underset{n = 1}{\sum^{997}} (\tan^{2} x_{n} + \frac{1}{\tan^{2} x_{n}})$ $= \underset{n = 1}{\sum^{997}} (\tan^{2} x_{n} + \frac{1}{\tan^{2} x_{n}} - 2 + 2)$ $= \underset{n = 1}{\sum^{997}} {{(\tan x_{n} - \frac{1}{\tan x_{n}})}^{2} + 2}$ $⩾ \underset{n = 1}{\sum^{997}} {2} = 997 \times 2 = 1994$ $= i s v a l i d o n l y i f \tan x_{n} - \frac{1}{\tan x_{n}} = 0, i . e .$ $\tan x_{n} = \frac{1}{\tan x_{n}} \Rightarrow \tan x_{n} = \pm 1 \Rightarrow x_{n} = k_{n} π \pm \frac{π}{4}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum