Question
Evalute the following integrals:
$\int\big\{1+\tan\text{x}\tan(\text{x}+\theta)\big\}\text{dx}$

Answer

Since,
$\tan(\text{A}-\text{B})=\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}$
$\therefore\tan(\text{x}+\theta-\text{x})=\frac{\tan(\text{x}+\theta)-\tan\text{x}}{1+\tan(\text{x}+\theta)\tan\text{x}}$
$\Rightarrow 1+\tan(\text{x}+\theta)\tan\text{x}=\frac{\tan(\text{x}+\theta)-\tan\text{x}}{\tan\theta}$
$\Rightarrow\int1+\tan(\text{x}+\theta)\tan\text{x dx}$
$=\frac{1}{\tan\theta}\big[\int\tan(\text{x}+\theta)\text{dx}-\int\tan\text{x dx}\big]$
$=\frac{1}{\tan\theta}\big[-\log|\cos(\text{x}+\theta)++\log|\cos\text{x}|\big]+\text{C}$
$=\frac{1}{\tan\theta}\big[\log|\cos\text{x}|-\log|\cos(\text{x}+\theta)|\big]+\text{C}$
$=\frac{1}{\tan\theta}\log\Big|\frac{\cos\text{x}}{\cos(\text{x}+\theta)}\Big|+\text{C}$

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