Question

can u solve questions 20 (star)

Answer 1

The answer is 2cosec2x

look the attatchment for solution

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Answer 2

Answer:

Option(2) - 2cosec2x

Explanation:

Given y = log tanx.

Then,

$\frac{d}{dx} (log (tanx))$

Note: We know that d/dx(log x) = 1/x

$=\ \textgreater \ \frac{1}{tanx} * \frac{d}{dx} (tan(x))$

Note: We know that d/dx(tanx) = sec^2x

$= \ \textgreater \ \frac{1}{tanx} * sec^2x$

$= \ \textgreater \ \frac{1}{ \frac{sinx}{cosx} } * \frac{1}{cos^2x}$

$= \ \textgreater \ \frac{cosx}{sinx} * \frac{1}{cosx * cosx}$

$= \ \textgreater \ \frac{1}{cosxsinx}$

$= \ \textgreater \ \frac{1}{ \frac{1}{2}sin(2x) }$

$= \ \textgreater \ \frac{1}{ \frac{sin(2x)}{2} }$

$= \ \textgreater \ \frac{2}{sin(2x)}$

2cosec2x.


Hope this helps!