Grade 12Differential Calculus

Prove that 2 tan^-1(1/5) + tan^-1(1/8) = tan^-1(4/7)

taniska

11 Years agoGrade 12

1 Answer

Jitender Singh

11 Years ago

Ans:
Hello Student,
Please find answer to your question below

$A = 2tan^{-1}\frac{1}{5} + tan^{-1}\frac{1}{8}$
$A = tan^{-1}\frac{1}{5}+tan^{-1}\frac{1}{5} + tan^{-1}\frac{1}{8}$
$tan^{-1}x + tan^{-1}y = tan^{-1}(\frac{x+y}{1-xy})$
$A = tan^{-1}\frac{1}{5}+(tan^{-1}\frac{1}{5} + tan^{-1}\frac{1}{8})$
$A = tan^{-1}\frac{1}{5}+(tan^{-1}\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}.\frac{1}{8}})$
$A = tan^{-1}\frac{1}{5}+(tan^{-1}\frac{\frac{13}{40}}{\frac{39}{40}})$
$A = tan^{-1}\frac{1}{5}+(tan^{-1}\frac{1}{3})$
$A = (tan^{-1}\frac{1}{5}+tan^{-1}\frac{1}{3})$
$A = (tan^{-1}\frac{\frac{1}{5}+\frac{1}{3}}{1-\frac{1}{5}.\frac{1}{3}}})$
$A = (tan^{-1}\frac{\frac{8}{15}}{\frac{14}{15}})$
$A = (tan^{-1}\frac{8}{14})$
$A = tan^{-1}\frac{4}{7}$