NCERT Class XI Mathematics - Limits and Derivatives - Solutions
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Question : 58
Total: 72
cosec x cot x
Solution:
Let f(x) = cosec x cot x
⇒ f (x) =
.
⇒ f (x) =c o s x ( s i n x ) − 2
Differentiating (i) with respect to x, we get
( s i n x ) − 2 ( s i n x ) − 3
= –( s i n x ) – 1 ( s i n x ) – 3 ( s i n x ) − 1 −
= -( s i n x ) − 1 2 c o t 2 x
= - cosex c - 2c o t 2
= – cosec x [1 + 2c o t 2
= – cosec x [1 +c o t 2 x + c o t 2 x
∴
[ c o s e c 2 x + c o t 2 x ] c o s e c 3 x − c o s e c x . c o t 2 x
⇒ f (x) =
⇒ f (x) =
Differentiating (i) with respect to x, we get
= –
= -
= - cosex c - 2
= – cosec x [1 + 2
= – cosec x [1 +
∴
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