Question

someone pls help me out with this....and DON'T SPAM!

Answer 1

Given :

$\: \tt \: y \: = \frac{1}{2} \: log \bigg( \frac{1 + cos \: 2x}{1 - cos \: 2x} \bigg)$

To Prove :

$\tt \: \frac{dy}{dx} = 2 \: cosec \: x$

Proof :

$\dashrightarrow \sf \: y \: = \frac{1}{2} \: log \bigg( \frac{1 - cos \: 2x}{1 + cos \: 2x} \bigg) \\ \\ \\ \underline{ \bf{ \pink{we \: have \: }}} \\ \\ \\ \underline{ \boxed{ \pmb{ \blue{1 + cos(2 \theta) = 2co {s}^{2} \theta}}}} \bigstar \\ \\ \\ \underline{ \boxed{ \pmb{ \blue{1 + cos(2 \theta) = 2{sin}^{2} \theta}}}} \bigstar \\ \\ \\ \dashrightarrow \sf \:y \: = \frac{1}{2} \: log \bigg( \frac{ {2 \: sin}^{2}x }{ {2 \: cos}^{2} x} \bigg) \\ \\ \\ \dashrightarrow \sf \:y = \frac{1}{2} \: log \: {(tan \: x)}^{2} \\ \\ \\ \dashrightarrow \sf \:y = \frac{1}{2} \times 2 \: log(tan \: x) \\ \\ \\ { \underline{ \boxed{ \bf{ \blue{ \because \: log {(a}^{m}) = m \: \times log(a)}}}}} \bigstar \\ \\ \\ \dashrightarrow \sf \:y = log \: (tan \: x)$

On differentiating y with respect to x we get ,

$\tt { \purple{ \frac{dy}{dx} = \frac{d}{dx}(log(tan \: x))}}$

We know that,

$\tt{ \purple{ \frac{d}{dx} (log \: x) = \frac{1}{x}}}$

$\dashrightarrow \sf \: \frac{dy}{dx} \: = \frac{1}{tan \: x} \: \: \frac{d}{dx} \: (tan \: x) \\ \\ \\ \bf \underline{ \red{using \: \: chain \: \: rule : }}$

$\\ \\ \\ \dashrightarrow \sf \: \frac{d}{dx} (tan \: x) = {sec }^{2} x \\ \\ \\ \dashrightarrow \sf \: \frac{dy}{dx} = \frac{1}{tan \: x} \times {sec}^{2} x$

$\\ \\ \\ \dashrightarrow \sf \: \frac{dy}{dx} = \frac{se {c}^{2} x}{tan \: x} \\ \\ \\ \dashrightarrow \sf \: \frac{dy}{dx} \: = \frac{( \frac{1}{ {cos}^{2} x}) }{ (\frac{sin \: x}{cos \: x} )} \\ \\ \\ \dashrightarrow \sf \: \frac{dy}{dx} \: = \frac{1}{ {cos}^{2} x} \times \frac{cos \: x}{sin \: x}$

$\dashrightarrow \sf \: \frac{dy}{dx} = \frac{1}{sin \: x \: cos \: x} \\ \\ \\ \underline{ \bf{ \red{we \: have \: sin(2 \theta) = 2sin \theta \: cos \theta}}}$

$\dashrightarrow \sf \: \frac{dy}{dx} = \frac{1}{ (\frac{sin \: 2x}{2}) } \\ \\ \\ \dashrightarrow \sf \: \frac{dy}{dx} \: \: = \frac{2}{sin \: 2x} \\ \\ \\ \dashrightarrow \sf \: \frac{dy}{dx} = 2cosec \: 2x$

${ \underline{ \boxed{ \rm{ \orange{ \therefore\frac{dy}{dx} = 2 \: cosec \: 2x}}}}}$

Hence Proved !!!

Answer 2

Step-by-step explanation:

$Given :</p><p></p><p>\begin{gathered} \: \tt \: y \: = \frac{1}{2} \: log \bigg( \frac{1 + cos \: 2x}{1 - cos \: 2x} \bigg) \\ \end{gathered}y=21log(1−cos2x1+cos2x)</p><p></p><p>To Prove :</p><p></p><p>\begin{gathered} \tt \: \frac{dy}{dx} = 2 \: cosec \: x \\ \end{gathered}dxdy=2cosecx</p><p></p><p>Proof :</p><p>[tex]\begin{gathered} \dashrightarrow \sf \: y \: = \frac{1}{2} \: log \bigg( \frac{1 - cos \: 2x}{1 + cos \: 2x} \bigg) \\ \\ \\ \underline{ \bf{ \pink{we \: have \: }}} \\ \\ \\ \underline{ \boxed{ \pmb{ \blue{1 + cos(2 \theta) = 2co {s}^{2} \theta}}}} \bigstar \\ \\ \\ \underline{ \boxed{ \pmb{ \blue{1 + cos(2 \theta) = 2{sin}^{2} \theta}}}} \bigstar \\ \\ \\ \dashrightarrow \sf \:y \: = \frac{1}{2} \: log \bigg( \frac{ {2 \: sin}^{2}x }{ {2 \: cos}^{2} x} \bigg) \\ \\ \\ \dashrightarrow \sf \:y = \frac{1}{2} \: log \: {(tan \: x)}^{2} \\ \\ \\ \dashrightarrow \sf \:y = \frac{1}{2} \times 2 \: log(tan \: x) \\ \\ \\ { \underline{ \boxed{ \bf{ \blue{ \because \: log {(a}^{m}) = m \: \times log(a)}}}}} \bigstar \\ \\ \\ \dashrightarrow \sf \:y = log \: (tan \: x)\end{gathered}⇢y=21log(1+cos2x1−cos2x)wehave1+cos(2θ)=2cos2θ1+cos(2θ)=2cos2θ★1+cos(2θ)=2sin2θ1+cos(2θ)=2sin2θ★⇢y=21log(2cos2x2sin2x)⇢y=21log(tanx)2⇢y=21×2log(tanx)∵log(am)=m×log(a)★⇢y=log(tanx)$