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 108780 by 150505R last updated on 19/Aug/20

Answered by bobhans last updated on 19/Aug/20

cot^2 81° = tan^2 9 cot^2 63° = tan^2 27 (∗) (1/(2−cot^2 9°)) + (1/(2−tan^2 9°)) + (1/(2−cot^2 27°))+(1/(2−tan^2 27°)) = μ let 9° = x μ= (1/(2−tan^2 x))+(1/(2−(1/(tan^2 x))))+(1/(2−(1/(tan^2 3x))))+(1/(2−tan^2 3x))

$$\mathrm{cot}\:^{\mathrm{2}} \:\mathrm{81}°\:=\:\mathrm{tan}\:^{\mathrm{2}} \:\mathrm{9}\: \\ $$$$\mathrm{cot}\:^{\mathrm{2}} \:\mathrm{63}°\:=\:\mathrm{tan}\:^{\mathrm{2}} \:\mathrm{27}\: \\ $$$$\left(\ast\right)\:\frac{\mathrm{1}}{\mathrm{2}−\mathrm{cot}\:^{\mathrm{2}} \:\mathrm{9}°}\:+\:\frac{\mathrm{1}}{\mathrm{2}−\mathrm{tan}\:^{\mathrm{2}} \:\mathrm{9}°}\:+\:\frac{\mathrm{1}}{\mathrm{2}−\mathrm{cot}\:^{\mathrm{2}} \:\mathrm{27}°}+\frac{\mathrm{1}}{\mathrm{2}−\mathrm{tan}\:^{\mathrm{2}} \:\mathrm{27}°}\:=\:\mu \\ $$$${let}\:\mathrm{9}°\:=\:{x}\: \\ $$$$\mu=\:\frac{\mathrm{1}}{\mathrm{2}−\mathrm{tan}\:^{\mathrm{2}} {x}}+\frac{\mathrm{1}}{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{tan}\:^{\mathrm{2}} {x}}}+\frac{\mathrm{1}}{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{tan}\:^{\mathrm{2}} \mathrm{3}{x}}}+\frac{\mathrm{1}}{\mathrm{2}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{3}{x}} \\ $$

Commented by bobhans last updated on 19/Aug/20

let tan x = q

$${let}\:\mathrm{tan}\:{x}\:=\:{q}\: \\ $$$$ \\ $$

Commented by 150505R last updated on 19/Aug/20

can i solve it further ?

$${can}\:{i}\:{solve}\:{it}\:{further}\:? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com