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Question-108780




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

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