Question

To prove that tan ^-1x + tan^-1 2x/1-x² = tan^-1 (3x-x³/1-3x²)​

Answer 1

Answer:

tex]\huge\tt {sec}^{2} x = \frac{dt}{dx}[/tex]

⇛$\huge\tt{dx \frac{dt}{ {sec}^{2}x }}$

ㅤ ㅤ ㅤ ㅤ ㅤ

⇛$\huge\tt∴∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx$

⇛$\huge\tt ∫ {(t)}^{ - \frac{1}{2} } \times {sec}^{2} x \times \frac{dt}{ {sec}^{2}x }$

⇛$\huge\tt ∫ {t}^{ - \frac{1}{2} }$ㅤ ㅤ

⇛ $\huge\tt\frac{ {t}^{ - \frac{1}{2} + 1} }{ - \frac{1}{2} + 1 }$

⇛ $\huge\tt \frac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} } + c = 2 {t}^{ \frac{1}{2} } + c = 2 \sqrt{t}$

⇛$\huge2 \sqrt{t} + c = 2 \sqrt{tanx}$

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Answer 2

Answer:

Hence proved,,

Step-by-step explanation:

for detailed explanation of the question check attachment,,,

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