Prove the following
Question:

If $\cos (\alpha+\beta)=\frac{4}{5}$ and $\sin (\alpha-\beta)=\frac{5}{13}$, where α lie between 0 and π/4, find value of tan 2α

[Hint: Express tan 2α as tan (α + β + α – β]

Solution:

According to the question,

cos(α + β) = 4/5 …(i)

We know that,

sin x = √(1 – cos2x)

Therefore,

sin (α + β) = √(1 – cos2(α + β))

⇒ sin (α + β) = √(1 – (4/5)2) = 3/5 …(ii)

Also,

sin(α – β) = 5/13 {given} …(iii)

we know that,

cos x = √(1 – sin2x)

Therefore,

cos (α – β) = √(1 – sin2(α – β))

⇒ cos (α – β) = √(1 – (5/13)2) = 12/13 …(iv)

Therefore,

tan 2α = tan (α + β + α – β)

We know that,

$\tan (\mathrm{x}+\mathrm{y})=\frac{\tan \mathrm{x}+\tan \mathrm{y}}{1-\tan \mathrm{x} \tan y}$

$\therefore \tan 2 \alpha=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \tan (\alpha-\beta)}$

$\Rightarrow \tan 2 \alpha=\frac{\frac{\frac{s \cos (\alpha+\beta)}{\cos (\alpha+\beta)}+\frac{\sin (\alpha-\beta)}{\cos (\alpha-\beta)}}{1-\frac{\sin (\alpha+\beta)}{\cos (\alpha+\beta)} \times \frac{\sin (\alpha-\beta)}{\cos (\alpha-\beta)}}}{1}$

From equation i, ii, iii and iv we have,

$\Rightarrow \tan 2 \alpha=\frac{\frac{\frac{3}{5}}{\frac{5}{5}}+\frac{\frac{5}{13}}{\frac{12}{13}}}{1-\frac{\frac{3}{5}}{\frac{4}{5}} \times \frac{\frac{5}{13}}{\frac{12}{13}}}$

$=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \times \frac{5}{12}}$

$=\frac{\frac{9+5}{12}}{1-\frac{15}{48}}$

$\Rightarrow \tan 2 \alpha=\frac{14}{12\left(\frac{33}{48}\right)}$

$\Rightarrow \tan 2 \alpha=\frac{14}{12\left(\frac{33}{48}\right)}$

$=\frac{56}{33}$

Hence, tan 2α = 56/33