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If $\tan (A+B)=p$ and $\tan (A-B)=q$, then show that $\tan 2 A$ $=\frac{p+q}{1-p q}$
Solution:
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Verified Answer
We have $\tan (A+B)=p$
and $\tan (A-B) \quad=q$
$\therefore \tan 2 A=\tan (A+B+A-B)$
$$
=\frac{\tan (A+B)+\tan (A-B)}{1-\tan (A+B) \tan (A-B)}
$$
$$
\left[\because \tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}\right]
$$
From Eqns. (i) and (ii)
$$
\tan 2 \mathrm{~A}=\frac{p+q}{1-p q}
$$
and $\tan (A-B) \quad=q$
$\therefore \tan 2 A=\tan (A+B+A-B)$
$$
=\frac{\tan (A+B)+\tan (A-B)}{1-\tan (A+B) \tan (A-B)}
$$
$$
\left[\because \tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}\right]
$$
From Eqns. (i) and (ii)
$$
\tan 2 \mathrm{~A}=\frac{p+q}{1-p q}
$$
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