Question

Prove that $\frac{\tan (45°+A)-\tan (45°-A)}{\tan (45°+A)+\tan (45°-A)}=\sin 2A$

Answer 1

Determine the proof of the given expression$\frac{\tan (45°+A)-\tan (45°-A)}{\tan (45°+A)+\tan (45°-A)}=\sin 2A$

Using formula:

$$\begin{gathered} \tan (A+B)=\frac{\tan A+\tan B}{1–\tan A\tan B} \\ \tan (A–B)=\frac{\tan A-\tan B}{1+\tan A\tan B} \end{gathered}$$

Solve the L.H.S part:

$$\begin{gathered} \frac{\tan (45°+A)-\tan (45°-A)}{\tan (45°+A)+\tan (45°-A)}=\frac{\frac{\tan 45°+tanA}{1–tan45°tanA}-\left(\frac{tan45°-tanA}{1+tan45°tanA}\right)}{\frac{\tan 45°+tanA}{1–tan45°tanA}+\left(\frac{tan45°-tanA}{1+tan45°tanA}\right)}\mathbf{\because }\mathbf{}\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathbf{45}\mathbf{°}\mathbf{}=1 \\ \Rightarrow =\frac{\frac{1+\tan A}{1–1\tan A}-\left(\frac{1-\tan A}{1+1\tan A}\right)}{\frac{1+\tan A}{1–1\tan A}+\left(\frac{1-\tan A}{1+1\tan A}\right)} \\ \Rightarrow =\frac{\frac{1+\tan A}{1–\tan A}-\frac{1-\tan A}{1+\tan A}}{\frac{1+\tan A}{1–\tan A}+\frac{1-\tan A}{1+\tan A}} \\ \Rightarrow =\frac{\frac{{(1+\tan A)}^2-{(1-\tan A)}^2}{(1–\tan A)(1+\tan A)}}{\frac{{(1+\tan A)}^2+{(1-\tan A)}^2}{(1–\tan A)(1+\tan A)}}\mathbf{\because }\mathbf{}{\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{)}}^{\mathbf{2}}\mathbf{}=a^2+b^2+2.a.b\mathbf{}\mathit{a}\mathit{n}\mathit{d}{\mathbf{(}\mathit{a}\mathbf{-}\mathit{b}\mathbf{)}}^{\mathbf{2}}\mathbf{}=a^2+b^2-2.a.b \\ \Rightarrow =\frac{\frac{1+{\tan }^{2}A+2\tan A-(1+{\tan }^{2}A-2\tan A)}{(1–{\tan }^{2}A)}}{\frac{(1+{\tan }^{2}A+2\tan A)+1+{\tan }^{2}A-2\tan A}{(1–{\tan }^{2}A)}} \\ \Rightarrow =\frac{\frac{1+{\tan }^{2}A+2\tan A-1-{\tan }^{2}A+2\tan A}{(1–{\tan }^{2}A)}}{\frac{(1+{\tan }^{2}A+2\tan A)+1+{\tan }^{2}A-2\tan A}{(1–{\tan }^{2}A)}} \\ \Rightarrow =\frac{4\tan A}{(1–{\tan }^{2}A)}\times \frac{(1–{\tan }^{2}A)}{2+2{\tan }^{2}A} \\ \Rightarrow =\frac{4\tan A}{2+2{\tan }^{2}A} \\ \Rightarrow =\frac{4\tan A}{2(1+{\tan }^{2}A)}\mathbf{\because }\mathbf{}\mathbf{1}\mathbf{+}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{2}}\mathbf{}\mathit{x}\mathbf{}\mathbf{=}se{c}^{2}x \\ \Rightarrow =\frac{2\tan A}{se{c}^{2}A} \\ \Rightarrow =\frac{2{\displaystyle \frac{sinA}{\cos A}}}{{\displaystyle \frac{1}{{\cos }^{2}A}}} \\ \Rightarrow =2\frac{sinA}{\cos A}\times \frac{{\cos }^2\mathrm{A}}{1} \\ \Rightarrow =2sinA\cos A \\ \Rightarrow =sin2A \end{gathered}$$

Hence, the L.H.S = R.H.S.