Question

The value of ${lim}_{x\to 0}\frac{tan(a+x)-tan(a-x)}{ta{n}^{-1}(a+x)-ta{n}^{-1}(a-x)}$ is equal to

• A. $\frac{(1+a^2)}{si{n}^{2}a}$
• B. $\frac{(1+a^2)}{co{s}^{2}a}$
• C. $\frac{(1-a^2)}{co{s}^{2}a}$
• D. $\frac{a^2-1}{si{n}^{2}a}$

Answer 1

Answer: (B)

$\frac{(1+a^2)}{co{s}^{2}a}$

$\underset{x\to 0}{lim}\frac{tan(a+x)-tan(a-x)}{ta{n}^{-1}(a+x)-ta{n}^{-1}(a-x)}$

when we substitute x = 0 we get $\frac{0}{0}$ form. Thus, we can apply L-Hospitals

rule i.e can differentiate numerator and denominator

separately to get same limit. Thus,

$\underset{x\to 0}{lim}\frac{tan(a+x)-tan(a-x)}{ta{n}^{-1}(a+x)-ta{n}^{-1}(a-x)}=\underset{x\to 0}{lim}\frac{se{c}^2(a+x)-se{c}^2(a-x).(-1)}{\frac{1}{1+(a+x{)}^2}-\frac{1}{1+(a-x{)}^2}(-1)}$

$\underset{x\to 0}{lim}\frac{se{c}^2(a+x)+se{c}^2(a-x)}{\frac{1}{1+(a+x{)}^2}+\frac{1}{1+(a-x{)}^2}}$

now, can substiute x = 0

$=\frac{se{c}^{2}a+se{c}^{2}a}{\frac{1}{1+a^2}+\frac{1}{1+a^2}}$

$=se{c}^{2}a=(1+a^2)$

$=\frac{1+a^2}{co{s}^{2}a}$