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Question Number 82929 by VBash last updated on 26/Feb/20

$$if2B+A=45°$$$$showthat;$$$$tanB=\frac{1-2tanA-{tan}^{2}A}{1+2tanA-{tan}^{2}A}$$

Answered by jagoll last updated on 26/Feb/20

$$\mathrm{A}+\mathrm{B}={45}^{\mathrm{o}}-\mathrm{B}$$$$\tan (\mathrm{A}+\mathrm{B})=\tan ({45}^{\mathrm{o}}-\mathrm{B})$$$$\frac{\tan \mathrm{A}+\tan \mathrm{B}}{1-\tan \mathrm{A}.\tan \mathrm{B}}=\frac{1-\tan \mathrm{B}}{1+\tan \mathrm{B}}$$$$(\tan \mathrm{A}+\tan \mathrm{B})(1+\tan \mathrm{B})=$$$$(1-\mathrm{tanA}\tan \mathrm{B})(1-\tan \mathrm{B})$$$$\tan \mathrm{A}=\frac{\tan {}^2\mathrm{B}+2\tan \mathrm{B}-1}{\tan {}^2\mathrm{B}-2\tan \mathrm{B}-1}$$$$\mathrm{or}\tan \mathrm{A}=\frac{1-2\tan \mathrm{B}-\tan {}^2\mathrm{B}}{1+2\tan \mathrm{B}-\tan {}^2\mathrm{B}}$$$$$$

Commented by john santu last updated on 26/Feb/20

$$\mathrm{good}\mathrm{sir}$$

Commented by som(math1967) last updated on 26/Feb/20

$$Howtan(45°-B)=\frac{1-tanA}{1+tanA}???$$

Commented by jagoll last updated on 26/Feb/20

$$\mathrm{hahaha}..\mathrm{sorry}.\mathrm{it}\mathrm{my}\mathrm{typo}$$

Answered by TANMAY PANACEA last updated on 26/Feb/20

$$tan\left(2B\right)=tan(45-A)$$$$lettanA=atanB=b$$$$\frac{2b}{1-b^2}=\frac{1-a}{1+a}$$$$\frac{2b+1-b^2}{2b-1+b^2}=\frac{1-a+1+a}{1-a-1-a}$$$$\frac{2b+1-b^2}{2b-1+b^2}=\frac{2}{-2a}$$$$\frac{-a}{1}=\frac{2b-1+b^2}{2b+1-b^2}$$$$a=\frac{b^2+2b-1}{b^2-2b-1}plsrecheckthequestion$$$$$$$$$$$$$$$$$$$$$$$$$$

Commented by jagoll last updated on 26/Feb/20

$$\mathrm{it}\mathrm{that}\mathrm{the}\mathrm{question}\mathrm{wrong}$$

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