Question

write in simplest form
tan^-1 √1+x^2 - 1 / x​

Answer 1

Question : -

Write in simplest form

$\bf \: {tan}^{ - 1} \dfrac{ \sqrt{1 + {x}^{2} } - 1}{x}$

Answer :-

Given :-

$\bf \: {tan}^{ - 1} \dfrac{ \sqrt{1 + {x}^{2} } - 1}{x}$

To find :-

• Simplest form

Solution :-

$\bf \:Consider \: {tan}^{ - 1} \dfrac{ \sqrt{1 + {x}^{2} } - 1}{x}$

Put x = tanθ, we get

$\bf\implies \:{tan}^{ - 1} \dfrac{ \sqrt{1 + {tan}^{2}θ } - 1}{tanθ}$

$\bf\implies \:{tan}^{ - 1} \dfrac{ \sqrt{{sec}^{2} θ} - 1}{tanθ}$

$\bf\implies \: {tan}^{ - 1} (\dfrac{secθ - 1}{tanθ} )$

$\bf\implies \: {tan}^{ - 1}( \dfrac{\dfrac{1}{cosθ} - 1}{\dfrac{sinθ}{cosθ} } )$

$\bf\implies \: {tan}^{ - 1} (\dfrac{1 - cosθ}{sinθ} )$

$\bf\implies \: {tan}^{ - 1} (\dfrac{2 {sin}^{2} \dfrac{θ}{2} }{2sin\dfrac{θ}{2} cos\dfrac{θ}{2} } )$

$\bf\implies \: {tan}^{ - 1} (tan\dfrac{θ}{2} )$

$\bf\implies \:\dfrac{1}{2} θ$

$\bf\implies \:\dfrac{1}{2} {tan}^{ - 1} x$