Question

Simplify the function: tan⁻¹$[x + \sqrt{1 + x^{2}}\ ]$; x ∈ R.

Answer 1

Answer:

$tan^{-1}[x+\sqrt{1+x^{2}}]=\frac{\pi}{4}+\frac{1}{2}tan^{-1}x$

Step-by-step explanation:

to simplify given function,we must eliminate inverse trigonometry function

as we know that

1+tan²x=sec²x

so put x= tan θ

tan⁻¹[ tan θ+√(1+tan²θ)]

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

as we know that

$sin^{2}\:\theta+cos^{2}\:\theta=1\\cos\:2\theta=cos^{2}\:\theta-sin^{2}\:\theta\\\\sin2\theta=2sin\:\theta\:cos\:\theta$

$tan^{-1}(\frac{sin^{2}\:\frac{\theta}{2}+cos^{2}\:\frac{\theta}{2}+2sin\:\frac{\theta}{2}cos\:\frac{\theta}{2}}{cos^{2}\:\frac{\theta}{2}-sin^{2}\:\frac{\theta}{2}})$

$tan^{-1}(\frac{(sin\frac{\theta}{2}+cos\frac{\theta}{2})^{2} }{(cos\frac{\theta}{2}+sin\frac{\theta}{2})(cos\frac{\theta}{2}-sin\frac{\theta}{2})} )\\\\\\=tan^{-1}(\frac{(sin\frac{\theta}{2}+cos\frac{\theta}{2}) }{(cos\frac{\theta}{2}-sin\frac{\theta}{2})} )\\\\on\:\:divide\:\:by\:cos\frac{\theta}{2}\\\\=tan^{-1}(\frac{(tan\frac{\theta}{2}+1) }{(1-tan\frac{\theta}{2})} )\\\\\\or\\\\\\\\=tan^{-1}(\frac{(1+tan\frac{\theta}{2}) }{(1-tan\frac{\theta}{2})} )$

as we know that tan45°=1

$tan^{-1}(\frac{(tan\frac{\pi}{4}+tan\frac{\theta}{2}) }{(1-tan\frac{\pi}{4}tan\frac{\theta}{2})}\\\\\\=tan^{-1}[tan(\frac{\pi}{4}+\frac{\theta}{2})]\\\\=\frac{\pi}{4}+\frac{\theta}{2}$

now place the value of θ =tan⁻¹x

$tan^{-1}[x+\sqrt{1+x^{2}}]=\frac{\pi}{4}+\frac{1}{2}tan^{-1}x$