Question

solve the following problem by taking x=cot theta only....
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Answer 1

Solution:

Given Function:

$= { \tan }^{ - 1} \frac{ \sqrt{1 + {x}^{2} - 1} }{x}$

$\textbf{Put} \: x = \tan∅⇒∅ = { \tan }^{ - 1} x$

$∴ { \tan}^{ - 1} \frac{ \sqrt{1 + {x}^{2} - 1 } }{x} = { \tan }^{ - 1} ( \frac{ \sqrt{1 + { \tan }^{2}∅ - 1 } }{ \tan∅} )$

$∴1 + { \tan }^{2} ∅ = { \sec }^{2}∅ ,$

$\textbf{Where} \: x \: \textbf{Tends \: to \: } \: \tan∅$

$= { \tan }^{ - 1} ( \frac{ 1 - \cos∅ }{ \sin ∅} )$

$= { \tan}^{ - 1} ( \frac{2 { \sin }^{2} \frac{∅}{2} }{2 \sin \: \frac{∅}{2} \cos \: \frac{∅}{2} } )$

$= { \tan}^{ - 1} ( \tan \: \frac{∅}{2} )$

$= \frac{∅}{2}$

$= \frac{1}{2} { \tan}^{ - 1} x$

$\\ \\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}$