Question

$\boxed { \sf [tex]\sf \large \:\underline{ \red{Question} }:-$ }[/tex]
$solve \: {tan}^{ - 1} ( \frac{x}{y} ) { - ten}^{ - 1} \: \frac{x - y}{x + y} \:$
is equal to
(a) π/2
(b) π/3
(c) π/4
(d) -3π/5​

Answer 1

Answer:

$\sf \large \:\underline{ \red{Given} }:-$

${ten}^{ - 1} \: (\frac{x}{y} ) \: - te {n}^{ - 1} \: \frac{x - y}{x + y}$

$= te {n}^{ - 1} [ \frac{ \frac{x}{y} - \frac{x - y}{x + y} }{1 + ( \frac{x}{y}) (\frac{x - y}{x + y}) } ]$

$❥ \: (ta {n}^{ - 1} y - { \tan}^{ - 1} \: y = {tan}^{1} - \frac{x - y}{x + y}$

$= {tan}^{ - 1} \frac{ \frac{ \frac{x(x + y) - (x - y)}{y(x + y)} }{y(x + y) + x(x - y)} }{y(x + y)}$

$= {tan}^{ - 1} \: \: (\frac{ {x}^{2} + xy - xy + {y}^{2} }{xy + {y}^{2} + {x}^{2} + xy } )$

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

$\red {hence \:,the \: correct \: answer \: is } \: \pink{c}$

Answer 2

Step-by-step explanation:

$\bf\huge{Question:-}$

$\boxed { \sf [tex]\sf \large \:\underline{ \red{Question} }:-$ }[/tex]

$solve \: {tan}^{ - 1} ( \frac{x}{y} ) { - ten}^{ - 1} \: \frac{x - y}{x + y} \:$

is equal to

(a) π/2

(b) π/3

(c) π/4

(d) -3π/5

$\bf\huge{answer:-}$

your answer is option C