Question

Prove that $tan^{-1}(\frac{1}{7}) + tan^{-1}(\frac{1}{8}) = cot^{-1}(\frac{201}{43}) + cot^{-1} 18$

Answer 1

hope it help you dear.......

Answer 2

Answer:

Step-by-step explanation:

As we know that

$tan^{-1}x+tan^{-1}y=tan^{-1}(\frac{x+y}{1-xy} )\\\\\\tan^{-1}\frac{1}{7} +tan^{-1}\frac{1}{8} =tan^{-1}(\frac{\frac{1}{7} +\frac{1}{8} }{1-\frac{1}{7} \frac{1}{8}} )\\\\\\=tan^{-1}(\frac{15}{55} )\\\\=tan^{-1}(\frac{3}{11} )$...eq1

as we know that

$cot^{-1}x=tan^{-1}(\frac{1}{x})\\ \\so\\\\ cot^{-1}\frac{201}{43} =tan^{-1}(\frac{43}{201})\\\\cot^{-1}18=tan^{-1}(\frac{1}{18})$

$tan^{-1}x+tan^{-1}y=tan^{-1}(\frac{x+y}{1-xy} )\\\\\\tan^{-1}\frac{43}{201} +tan^{-1}\frac{1}{18} =tan^{-1}(\frac{\frac{43}{201} +\frac{1}{18} }{1-\frac{43}{201} \frac{1}{18}} )\\\\\\=tan^{-1}(\frac{774+201}{3618-43} )\\\\=tan^{-1}(\frac{975}{3575} )\\\\\\==tan^{-1}(\frac{3}{11} )$...eq2

from eq1 and eq2

LHS=RHS

hence proved