prove-that-tan-2tan2-4tan-4-8cot8-cot-

Question Number 43467 by peter frank last updated on 11/Sep/18

p r o v e t h at \tan θ + 2 t a n 2 θ + 4 t a n 4 θ

+ 8 \cot 8 θ = c o t θ

Answered by ajfour last updated on 11/Sep/18

8 \cot 8 θ + 4 \tan 4 θ + 2 \tan 2 θ + \tan θ

= \frac{8 (1 - \tan^{2} 4 θ)}{2 \tan 4 θ} + 4 \tan 4 θ + \dots

= 4 \cot 4 θ + 2 \tan 2 θ + \tan θ

= \frac{4 (1 - \tan^{2} 2 θ)}{2 \tan 2 θ} + 2 \tan 2 θ + \tan θ

= 2 \cot 2 θ + \tan θ

= \frac{2 (1 - \tan^{2} θ)}{2 \tan θ} + \tan θ = \cot θ .

Commented by peter frank last updated on 11/Sep/18

t h a n k s

Commented by peter frank last updated on 11/Sep/18

sir the third line am not understood

how you get 4 \cot 4 θ

Commented by peter frank last updated on 11/Sep/18

sir the third line am not understood

how you get 4 \cot 4 θ

Commented by ajfour last updated on 11/Sep/18

\frac{8}{2 \tan 4 θ} = 4 \cot 4 θ

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Sep/18

t a n θ - c o t θ

= t a n θ - \frac{1}{t a n θ}

= \frac{{t a n}^{2} θ - 1}{t a n θ}

= \frac{- 2 (1 - {t a n}^{2} θ)}{2 t a n θ}

= - 2 c o t 2 θ

s o t a n θ = c o t θ - 2 c o t 2 θ

2 t a n 2 θ = 2 (c o t 2 θ - 2 c o t 4 θ)

4 t a n 4 θ = 4 (c o t 4 θ - 2 c o t 8 θ)

8 c o t 8 θ = 8 c o t 8 θ

n o w a d d t h e m R H S = c o t θ

n e w t r i c k s \dots n o n e e d t o c a l c u l a t d t a n 2 θ, t a n 4 θ

c o t 8 θ e t c \dots

Commented by peter frank last updated on 11/Sep/18

t h a n k s