Question

from differentiation -
$sin \: x. \: cosec \: x + sec \: x. \: cot \: x$
right answer will brainliest ✔✌​

Answer 1

Answer:

cosec x - sin x ] [ sec x - cos x ] [ tan x + cotx ] = 1

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= [ 1sinx - sin x ] [ 1cosx - cos x ] [ sinxcosx + cosxsinx ] = [ 1 - sin^2xsinx ] [ 1 - cos^2xcosx ] [ sin^2x + cos^2xcosxsinx ] = cos^2xsinx. sin^2xcosx

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Answer 2

Answer:

-cot(x) cosec(x)

Step-by-step explanation:

Consider the given function,

$\rm\sf f(x) = sin(x) \cdot cosec(x) + sec(x) \cdot cot(x)$

We know that:

$\boxed{\tt cosec(x) = \dfrac{1}{sin(x)}}$

$\boxed{\tt sec(x) = \dfrac{1}{cos(x)}}$

$\boxed{\tt cot(x) = \dfrac{cos(x)}{sin(x)}}$

Using these formulas, we get:

$\rm\sf f(x) = sin(x) \cdot\dfrac{1}{sin(x)} + \dfrac{1}{cos(x)} \cdot \dfrac{cos (x)}{sin(x)}$

$\rm\sf f(x) = 1 +\dfrac{1}{sin(x)}$

$\rm\sf f(x) = 1 +cosec(x)$

Now differentiating the function,

$\sf\implies f'(x) = \dfrac{d}{dx}(1 +cosec(x))$

$\sf\implies f'(x) = \dfrac{d}{dx}(1) +\dfrac{d}{dx}(cosec(x))$

Now using the  following formulas:

$\boxed{\tt \dfrac{d}{dx}(Costant) = 0}$

$\boxed{\tt \dfrac{d}{dx}( cosec(x)) = - cot(x) \cdot(cosec(x))}$

By using these results, we get:

$\sf\implies f'(x) = 0 - cot(x) \cdot cosec(x)$

$\sf\implies f'(x) = - cot(x) \cdot cosec(x)$

So the required answer is -cot(x) cosec(x).