Question

please please give me answer​

Answer 1

Step-by-step explanation:

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

here, put ax = tanΦ , Φ = tan^(-1)(ax)

$= \tan {}^{ - 1} (\frac{ \sqrt{1 + { \tan }^{2} \: theta } - 1}{ \tan \: theta } )$

we know, 1 + tan²Φ = sec²Φ

$= { \tan }^{ - 1} ( \frac{ \sqrt{ \sec ^{2} \: theta } - 1}{ \tan \: theta } )$

$= { \tan}^{ - 1} ( \frac{ \sec \: theta \: - 1 }{tan \: theta} )$

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

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

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

here we know ,

cos2Φ = 1 - 2sin²Φ

cos 2Φ -1 = -2sin²Φ

taking (-1) on both the sides

-cos 2Φ + 1 = 2sin²Φ

2sin²Φ = 1 - cos2Φ

1 - cosΦ = 2sin²(Φ/2)

And, Sin2Φ = 2sinΦ. cosΦ

therefor

sinΦ = 2 sin(Φ/2).cos (Φ/2)

$= { \tan}^{ - 1} ( \frac{2 { \sin}^{2} (\frac{theta}{2} )}{2 \sin( \frac{theta}{2} ) . \cos( \frac{theta}{2} ) })$

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

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

here, $theta = tan^{-1}(ax )$

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

$\boxed {= \frac{1}{2} tan^{-1}(ax)}$