Question

Simplify the function: tan⁻¹ (sec x + tan x).

Answer 1

Answer:

$tan^{-1}(sec\:x+tan\:x)=\frac{\pi}{4}+\frac{x}{2}$

Step-by-step explanation:

To simplify the given function

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

as we know that

$sin^{2}\:x+cos^{2}\:x=1\\cos\:2x=cos^{2}\:x-sin^{2}\:x\\\\sin2x=2sin\:x\:cos\:x$

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

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

as we know that tan45°=1

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

thus by this way given function can simplify.