find the number of values of cot θ where θ∈[(π/(12)) (π/2)] satisfying the equation [tan θ.[cot θ]]=1 ? (where [x] is greatest integer less than or equal to x)


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 147008 by gsk2684 last updated on 17/Jul/21

$f i n d t h e n u m b e r o f v a l u e s o f \cot θ$ $w h e r e θ \in [\frac{π}{12} \frac{π}{2}] s a t i s f y i n g t h e$ $e q u a t i o n [\tan θ . [\cot θ]] = 1 ?$ $(w h e r e [x] i s g r e a t e s t i n t e g e r$ $l e s s t h a n o r e q u a l t o x)$
Commented by gsk2684 last updated on 22/Jul/21
$[tan θ[cot θ]]=1⇒1≤tan θ[cot θ]<2 [cot θ]∈{0,1,2,3} if θ∈[(π/(12)) (π/2)] i)[cot θ]=1 ⇒1≤cot θ<2⇒cot^(−1) 2<θ≤(π/4) then (1/2)<tan θ≤1 &1≤tan θ<2 ⇒tan θ=1 ⇒θ=(π/4) ii)[cot θ]=2 ⇒2≤cot θ<3⇒cot^(−1) 3<θ≤cot^(−1) 2 then (1/3)<tan θ≤(1/2) &1≤2 tan θ<2 (1/3)<tan θ≤(1/2) &(1/2)≤ tan θ<1 ⇒tan θ=(1/2)⇒(π/(12))<θ<(π/4) iii)[cot θ]=3 ⇒3≤cot θ<cot (π/(12))⇒(π/(12))<θ≤cot^(−1) 3 then 2−(√3)<tan θ≤(1/3) &1≤3 tan θ<2 2−(√3)<tan θ≤(1/3) &(1/3)≤ tan θ<(2/3) ⇒tan θ=(1/3)⇒(π/(12))<θ<(π/4) ans 3 values exist$
$[\tan θ [\cot θ]] = 1 \Rightarrow 1 ⩽ \tan θ [\cot θ] < 2$ $[\cot θ] \in {0, 1, 2, 3} i f θ \in [\frac{π}{12} \frac{π}{2}]$ $i) [\cot θ] = 1$ $\Rightarrow 1 ⩽ \cot θ < 2 \Rightarrow \cot^{- 1} 2 < θ ⩽ \frac{π}{4}$ $t h e n \frac{1}{2} < \tan θ ⩽ 1 & 1 ⩽ \tan θ < 2$ $\Rightarrow \tan θ = 1$ $\Rightarrow θ = \frac{π}{4}$ $i i) [\cot θ] = 2$ $\Rightarrow 2 ⩽ \cot θ < 3 \Rightarrow \cot^{- 1} 3 < θ ⩽ \cot^{- 1} 2$ $t h e n \frac{1}{3} < \tan θ ⩽ \frac{1}{2} & 1 ⩽ 2 \tan θ < 2$ $\frac{1}{3} < \tan θ ⩽ \frac{1}{2} & \frac{1}{2} ⩽ \tan θ < 1$ $\Rightarrow \tan θ = \frac{1}{2} \Rightarrow \frac{π}{12} < θ < \frac{π}{4}$ $i i i) [\cot θ] = 3$ $\Rightarrow 3 ⩽ \cot θ < \cot \frac{π}{12} \Rightarrow \frac{π}{12} < θ ⩽ \cot^{- 1} 3$ $t h e n 2 - \sqrt{3} < \tan θ ⩽ \frac{1}{3} & 1 ⩽ 3 \tan θ < 2$ $2 - \sqrt{3} < \tan θ ⩽ \frac{1}{3} & \frac{1}{3} ⩽ \tan θ < \frac{2}{3}$ $\Rightarrow \tan θ = \frac{1}{3} \Rightarrow \frac{π}{12} < θ < \frac{π}{4}$ $a n s 3 v a l u e s e x i s t$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum