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Question Number 82767 by jagoll last updated on 24/Feb/20

If ((sin (A+θ))/(sin (B+θ))) = (√((sin 2A)/(sin 2B)))  prove that tan^2 θ = tan A.tan B

$$\mathrm{If}\:\frac{\mathrm{sin}\:\left(\mathrm{A}+\theta\right)}{\mathrm{sin}\:\left({B}+\theta\right)}\:=\:\sqrt{\frac{\mathrm{sin}\:\mathrm{2}{A}}{\mathrm{sin}\:\mathrm{2}{B}}} \\ $$$${prove}\:{that}\:\mathrm{tan}\:^{\mathrm{2}} \theta\:=\:\mathrm{tan}\:{A}.\mathrm{tan}\:{B} \\ $$

Commented by jagoll last updated on 24/Feb/20

thank you mr w and john

$${thank}\:{you}\:{mr}\:{w}\:{and}\:{john} \\ $$

Answered by mr W last updated on 24/Feb/20

((sin A cos θ+cos A sin θ)/(sin B cos θ+cos B sin θ))=(√((sin A cos A)/(sin B cos B)))  ((1+((tan θ)/(tan A)))/(1+((tan θ)/(tan B))))=((sin B)/(sin A))(√((sin A cos A)/( sin B cos B)))  ((1+((tan θ)/(tan A)))/(1+((tan θ)/(tan B))))=(√((tan B)/( tan A)))  1+((tan θ)/(tan A))=(√((tan B)/( tan A)))+((tan θ)/(√(tan A tan B)))  ((1/(tan A))−(1/(√(tan A tan B))))tan θ=(√((tan B)/( tan A)))−1  ((1/(√(tan A)))−(1/(√(tan B))))tan θ=(√(tan B))−(√(tan A))  ((tan θ)/(√(tan A tan B)))=1  tan θ=(√(tan A tan B))  ⇒tan^2  θ=tan A tan B

$$\frac{\mathrm{sin}\:{A}\:\mathrm{cos}\:\theta+\mathrm{cos}\:{A}\:\mathrm{sin}\:\theta}{\mathrm{sin}\:{B}\:\mathrm{cos}\:\theta+\mathrm{cos}\:{B}\:\mathrm{sin}\:\theta}=\sqrt{\frac{\mathrm{sin}\:{A}\:\mathrm{cos}\:{A}}{\mathrm{sin}\:{B}\:\mathrm{cos}\:{B}}} \\ $$$$\frac{\mathrm{1}+\frac{\mathrm{tan}\:\theta}{\mathrm{tan}\:{A}}}{\mathrm{1}+\frac{\mathrm{tan}\:\theta}{\mathrm{tan}\:{B}}}=\frac{\mathrm{sin}\:{B}}{\mathrm{sin}\:{A}}\sqrt{\frac{\mathrm{sin}\:{A}\:\mathrm{cos}\:{A}}{\:\mathrm{sin}\:{B}\:\mathrm{cos}\:{B}}} \\ $$$$\frac{\mathrm{1}+\frac{\mathrm{tan}\:\theta}{\mathrm{tan}\:{A}}}{\mathrm{1}+\frac{\mathrm{tan}\:\theta}{\mathrm{tan}\:{B}}}=\sqrt{\frac{\mathrm{tan}\:{B}}{\:\mathrm{tan}\:{A}}} \\ $$$$\mathrm{1}+\frac{\mathrm{tan}\:\theta}{\mathrm{tan}\:{A}}=\sqrt{\frac{\mathrm{tan}\:{B}}{\:\mathrm{tan}\:{A}}}+\frac{\mathrm{tan}\:\theta}{\sqrt{\mathrm{tan}\:{A}\:\mathrm{tan}\:{B}}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{tan}\:{A}}−\frac{\mathrm{1}}{\sqrt{\mathrm{tan}\:{A}\:\mathrm{tan}\:{B}}}\right)\mathrm{tan}\:\theta=\sqrt{\frac{\mathrm{tan}\:{B}}{\:\mathrm{tan}\:{A}}}−\mathrm{1} \\ $$$$\left(\frac{\mathrm{1}}{\sqrt{\mathrm{tan}\:{A}}}−\frac{\mathrm{1}}{\sqrt{\mathrm{tan}\:{B}}}\right)\mathrm{tan}\:\theta=\sqrt{\mathrm{tan}\:{B}}−\sqrt{\mathrm{tan}\:{A}} \\ $$$$\frac{\mathrm{tan}\:\theta}{\sqrt{\mathrm{tan}\:{A}\:\mathrm{tan}\:{B}}}=\mathrm{1} \\ $$$$\mathrm{tan}\:\theta=\sqrt{\mathrm{tan}\:{A}\:\mathrm{tan}\:{B}} \\ $$$$\Rightarrow\mathrm{tan}^{\mathrm{2}} \:\theta=\mathrm{tan}\:{A}\:\mathrm{tan}\:{B} \\ $$

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