find-the-number-of-values-of-cot-where-pi-12-pi-2-satisfying-the-equation-tan-cot-1-where-x-is-greatest-integer-less-than-or-equal-to-x-

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 \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

Facebook Tweet Pin