Question

بع
رمانسر
2.
و
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Answer 1

Explanation:

is it Urdu brother

Answer 2

Explanation:

Solution :

$\begin{gathered}\sf{First\:let\:us\:solve\:the\:identity\::\:\dfrac{2sin(x) - sin2(x)}{2sin(x) + sin2(x)}} \\ \\ \end{gathered}Firstletussolvetheidentity:2sin(x)+sin2(x)2sin(x)−sin2(x)</p><p>\begin{gathered}:\implies \sf{\dfrac{2sin(x) - sin2(x)}{2sin(x) + sin2(x)}} \\ \\ \end{gathered}:⟹2sin(x)+sin2(x)2sin(x)−sin2(x)</p><p>\begin{gathered}:\implies \sf{\dfrac{2sin(x) - 2sin(x)cos(x)}{2sin(x) + 2sin(x)cos(x)}} \:\:\:[\because \sf{sin2\theta = 2sin(\theta)cos(\theta)}] \\ \\ \\ \end{gathered}:⟹2sin(x)+2sin(x)cos(x)2sin(x)−2sin(x)cos(x)[∵sin2θ=2sin(θ)cos(θ)]</p><p>\begin{gathered}:\implies \sf{\dfrac{2sin(x)[1 - cos(x)]}{2sin(x)[1 + cos(x)]}} \\ \\ \\ \end{gathered}:⟹2sin(x)[1+cos(x)]2sin(x)[1−cos(x)]</p><p>\begin{gathered}:\implies \sf{\dfrac{1 - cos(x)}{1 + cos(x)}} \\ \\ \\ \end{gathered}:⟹1+cos(x)1−cos(x)</p><p>\begin{gathered}\textsf{Now, by using the identity of cos(x)\:} \\ \textsf{and substituting them in the equation, we get :} \\ \\ \bullet\:\sf{cos(x) = 2cos^{2}\bigg(\dfrac{x}{2} - 1\bigg)} \\ \\ \bullet\:\sf{cos(x) = 1 - 2sin^{2}\bigg(\dfrac{x}{2}\bigg)} \\ \\ \end{gathered}Now, by using the identity of cos(x)and substituting them in the equation, we get :∙cos(x)=2cos2(2x−1)∙cos(x)=1−2sin2(2x)</p><p>\begin{gathered}:\implies \sf{\dfrac{1 - 1 - 2sin^{2}\bigg(\dfrac{x}{2}\bigg)}{1 + 2cos^{2}\bigg(\dfrac{x}{2}\bigg) - 1}} \\ \\ \\ \end{gathered}:⟹1+2cos2(2x)−11−1−2sin2(2x)</p><p>\begin{gathered}:\implies \sf{\dfrac{2sin^{2}\bigg(\dfrac{x}{2}\bigg)}{2cos^{2}\bigg(\dfrac{x}{2}\bigg)}} \\ \\ \\ \end{gathered}:⟹2cos2(2x)2sin2(2x)</p><p>\begin{gathered}\textsf{Now, by using the identity of cos(x)\:} \\ \textsf{and substituting them in the equation, we get :} \\ \mapsto \sf{\dfrac{sin(x)}{cos(x)} = tan(x)} \\ \\ \end{gathered}Now, by using the identity of cos(x)and substituting them in the equation, we get :↦cos(x)sin(x)=tan(x)</p><p>\begin{gathered}:\implies \sf{tan^{2}\bigg(\dfrac{x}{2}\bigg)} \\ \\ \\ \end{gathered}:⟹tan2(2x)</p><p>\boxed{\therefore \sf{\dfrac{2sin(x) - sin2(x)}{2sin(x) + sin2(x)} = tan^{2}\bigg(\dfrac{x}{2}\bigg)}}∴2sin(x)+sin2(x)2sin(x)−sin2(x)=tan2(2x)</p><p>\begin{gathered}\textsf{We know that the value of\:}\sf{u\:is\: \dfrac{x}{2}}\:\textsf{So by substituting it in the equation, we get :} \\ \\ \end{gathered}We know that the value ofuis2xSo by substituting it in the equation, we get :</p><p>⠀⠀⠀⠀⠀⠀⠀⠀⠀\begin{gathered}:\implies \sf{tan^{2}\bigg(\dfrac{x}{2}\bigg)} \\ \\ \\ \end{gathered}:⟹tan2(2x)</p><p>⠀⠀⠀⠀⠀⠀⠀⠀⠀\begin{gathered}:\implies \sf{tan^{2}\bigg(\dfrac{x}{2}\bigg) = tan^{2}(u)} \\ \\ \\ \end{gathered}:⟹tan2(2x)=tan2(u)</p><p>⠀⠀⠀⠀⠀⠀⠀⠀⠀\begin{gathered}\boxed{\therefore \sf{tan^{2}\bigg(\dfrac{x}{2}\bigg) = tan^{2}(u)}} \\ \\ \\ \end{gathered}∴tan2(2x)=tan2(u)</p><p>\begin{gathered}\textsf{Now, by differentiating the function with respect to u, we get :-} \\ \\ \end{gathered}Now, by differentiating the function with respect to u, we get :-</p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{du} = \dfrac{d[tan^{2}u]}{du}} \\ \\ </p><p></p><p>[tex]\big\ {$