Question

$\sqrt{ \frac{tanx }{sinxcosx} }$​

Answer 1

$\orange{\bold{\underbrace{\overbrace{❥Question᎓}}}} \\ \huge\green\tt\frac{\sqrt{tanx} }{sinxcosx} \\ \huge\tt\frac{ \sqrt{tanx} }{sinxcosx} \\ \huge\tt \frac{ \sqrt{tanx} }{sinxcosx \times \frac{cosx}{cosx}}</p><p></p><p>ㅤ ㅤ ㅤ ㅤ ㅤ</p><p></p><p> \\ ⇛\huge\tt \frac{ \sqrt{tanx} }{sinx \times \frac{ {cos}^{2} x}{cosx}}ㅤ ㅤ \\ </p><p></p><p>⇛ \huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2} x \times \frac{sinx}{cosx} } \\ </p><p></p><p>ㅤ ㅤ⇛\huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2}x \times tanx } \\ </p><p>⇛\huge\tt {tan}^{ \frac{1}{2} - 1 } \times \frac{1}{ {cos}^{2} x} \\ </p><p></p><p>⇛\huge\tt {(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x⇛(tan)(tan)−21×cos2x1=(tanx)−21×sec2x⇛(tan) \\ </p><p></p><p>ㅤ ㅤ ㅤ ㅤ ㅤ</p><p></p><p>⇛\huge\tt {(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = ∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx⇛(tan)(tan)−21×cos2x1=∫(tanx)−21×sec2x×dx⇛(tan) \\ </p><p></p><p>ㅤ ㅤ ㅤ ㅤ ㅤ</p><p></p><p>\bold\blue{☛\: Let tanx=t}\\ </p><p></p><p>\bold\blue{☛ \:Differentiating \: both \: sides \: w.r.t.x}☛Differentiatingbothsidesw.r.t.x \\ </p><p></p><p>ㅤ ㅤ ㅤ ㅤ ㅤ</p><p></p><p>⇛\huge\tt {sec}^{2} x = \frac{dt}{dx}sec2x=dxdt</p><p> \\ </p><p>⇛\huge\tt{dx \frac{dt}{ {sec}^{2}x }}\\ </p><p></p><p>ㅤ ㅤ ㅤ ㅤ ㅤ</p><p></p><p>⇛\huge\tt∴∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx∴∫(tanx)−21×s \\ </p><p></p><p>⇛\huge\tt ∫ {(t)}^{ - \frac{1}{2} } \times {sec}^{2} x \times \frac{dt}{ {sec}^{2}x }∫(t)−21 \\ </p><p></p><p>⇛\huge\tt ∫ {t}^{ - \frac{1}{2} }∫t−21 ㅤ ㅤ \\ </p><p></p><p>⇛ \huge\tt\frac{ {t}^{ - \frac{1}{2} + 1} }{ - \frac{1}{2} + 1 }−21+1t−21+1 \\ </p><p></p><p>⇛ \huge\tt \frac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} } + c = 2 {t}^{ \frac{1}{2} } + c = 2 \sqrt{t}21t21+c=2t21+c=2t</p><p></p><p>⇛\huge2 \sqrt{t} + c = 2 \sqrt{tanx}2t+c=2tanx$

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Answer 2

Step-by-step explanation:

\sqrt{ \frac{tanx }{sinxcosx} }