Question

$y = {tan}^{ - 1} ( \frac{sinx}{1 + cosx} ) \\ \\ \: \: find \: \frac{dy}{dx}$
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Answer 1

$\huge \red{\mathfrak{solution}}$

Given:

$y = {tan}^{ - 1} ( \frac{sinx}{1 + cosx} )$

To find :

$\frac{dy}{dx}$

•First solve Y

$\rightarrow \: {tan}^{ - 1} (\frac{sinx}{1 + cosx} )$

As we know that :

$\boxed{ \bold{ \green{ \star \: sinx = 2sin \frac{x}{2} cos \frac{x}{2} }}} \\ \\ \boxed{ \bold{ \green{\star \:1 + cosx = 2 {cos}^{2} \frac{x}{2} }}}$

$\rightarrow \: {tan}^{ - 1} ( \frac{2sin \frac{x}{2} cos \frac{x}{2} }{2 {cos}^{2} \frac{x}{2} } ) \\ \\ \rightarrow {tan}^{ - 1} (tan \frac{x}{2} ) \\ \\ \rightarrow \: \frac{x}{2}$

Now differentiate w.r.t X

$\implies \: \frac{d}{dx} ( \frac{x}{2} ) \\ \\ \implies \frac{1}{2}$

Hence, the answer is

$\huge \boxed{ \bold{ \red {\frac{1}{2} }}}$

Answer 2

Question :---

• if y = Tan^(-1) [ sinx/(1+cosx) ] Find dy/dx ?

Formula used :---

• sinx = 2sinx/2 * cosx/2
• 1+cosx = 2cos²(x/2)
• sinx/cosx = tan x
• Tan^(-1) * TanA = A
• dy/dx of x = 1 .

Solution :--

sinx/(1+cosx) can be solved by Putting above told values ,

→ 2sinx/2 * cosx/2 / (2cos²(x/2))

( 2 cosx will cancel )

→ sin(x/2) / cos(x/2)

→ Tan(x/2)

putting the value in Question now,

→ y = Tan^(-1)(Tan(x/2)

→ y = x/2

So,

dy/dx = 1/2 [ d/dx (x) ]

→ dy/dx = 1/2 * 1

→ dy/dx = 1/2 (Ans)

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