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Question Number 137197 by Fikret last updated on 30/Mar/21

α , β ε (0 , (π/2))  tan^2 α = 1+2tan^2 β  ⇒(√2)cosα−cosβ=?

$$\alpha\:,\:\beta\:\epsilon\:\left(\mathrm{0}\:,\:\frac{\pi}{\mathrm{2}}\right) \\ $$$${tan}^{\mathrm{2}} \alpha\:=\:\mathrm{1}+\mathrm{2}{tan}^{\mathrm{2}} \beta\:\:\Rightarrow\sqrt{\mathrm{2}}{cos}\alpha−{cos}\beta=? \\ $$

Answered by MJS_new last updated on 31/Mar/21

cos x =(1/( (√(1+tan^2  x))))  let tan α =a∧tan β =b  a^2 =1+2b^2   ((√2)/( (√(1+a^2 ))))−(1/( (√(1+b^2 ))))=((√2)/( (√(2+2b^2 ))))−(1/( (√(1+b^2 ))))=0

$$\mathrm{cos}\:{x}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}} \\ $$$$\mathrm{let}\:\mathrm{tan}\:\alpha\:={a}\wedge\mathrm{tan}\:\beta\:={b} \\ $$$${a}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}+\mathrm{2}{b}^{\mathrm{2}} }}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}=\mathrm{0} \\ $$

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