Question

lim(x→0)$(\frac {1+tanx}{1+sinx})^{cosec x}$ = ?

• A. 0
• B. e
• C. 1
• D. $\frac {1}{e}$

Answer 1

Answer: (C)

1

Answer 2

Let's consider the limit as x approaches 0:
lim(x→0) $(\frac {1+tanx}{1+sinx})^{cosec x}$ as x approaches 0
Substituting x = 0 into the expression gives us:
$(\frac {1+tan(0)}{1+sin(0)})^{cosec (0)}$
Since tan(0) = 0 and sin(0) = 0, we have:
$(\frac {1+0}{1+0})$(cosec 0)
Simplifying further:
1(cosec 0)
1
Therefore, the correct option is (C) 1.