Question

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Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

We first memorize a formula on Trigonometric Inverse Function :

$\displaystyle \: { \tan}^{ - 1} a \: - { \tan}^{ - 1} b \: = { \tan}^{ - 1} \: \frac{a - b}{1 + ab}$

$where \: a \:, b \: > 0$

TO PROVE

$\displaystyle \: { \tan}^{ - 1} \: \frac{a - b}{1 + ab} + { \tan}^{ - 1} \: \frac{b - c}{1 + bc} + { \tan}^{ - 1} \: \frac{c - a}{1 + ca} = 0$

CALCULATION

Here

$\: a \:, b \: , \: c\: > 0$

Using the above mentioned formula

$\displaystyle \: { \tan}^{ - 1} \: \frac{a - b}{1 + ab} = { \tan}^{ - 1} a \: - { \tan}^{ - 1} b \:$

$\displaystyle \: { \tan}^{ - 1} \: \frac{b - c}{1 + bc} = { \tan}^{ - 1} b \: - { \tan}^{ - 1} c\:$

$\displaystyle \: { \tan}^{ - 1} \: \frac{c - a}{1 + ca} = { \tan}^{ - 1} c \: - { \tan}^{ - 1} a \:$

Hence

$\displaystyle \: { \tan}^{ - 1} \: \frac{a - b}{1 + ab} + { \tan}^{ - 1} \: \frac{b - c}{1 + bc} + { \tan}^{ - 1} \: \frac{c - a}{1 + ca}$

$= { \tan}^{ - 1} a \: - { \tan}^{ - 1} b \: + { \tan}^{ - 1} b \: - { \tan}^{ - 1} c \: + { \tan}^{ - 1} c \: - { \tan}^{ - 1} a \:$

$= 0$

Hence proved

Answer 2

Answer:

FORMULA TO BE IMPLEMENTED

We first memorize a formula on Trigonometric Inverse Function :

\displaystyle \: { \tan}^{ - 1} a \: - { \tan}^{ - 1} b \: = { \tan}^{ - 1} \: \frac{a - b}{1 + ab}tan

−1

a−tan

−1

b=tan

−1

1+ab

a−b

where \: a \:, b \: > 0wherea,b>0

TO PROVE

\displaystyle \: { \tan}^{ - 1} \: \frac{a - b}{1 + ab} + { \tan}^{ - 1} \: \frac{b - c}{1 + bc} + { \tan}^{ - 1} \: \frac{c - a}{1 + ca} = 0tan

−1

1+ab

a−b

+tan

−1

1+bc

b−c

+tan

−1

1+ca

c−a

=0

CALCULATION

Here

\: a \:, b \: , \: c\: > 0a,b,c>0

Using the above mentioned formula

\displaystyle \: { \tan}^{ - 1} \: \frac{a - b}{1 + ab} = { \tan}^{ - 1} a \: - { \tan}^{ - 1} b \:tan

−1

1+ab

a−b

=tan

−1

a−tan

−1

b

\displaystyle \: { \tan}^{ - 1} \: \frac{b - c}{1 + bc} = { \tan}^{ - 1} b \: - { \tan}^{ - 1} c\:tan

−1

1+bc

b−c

=tan

−1

b−tan

−1

c

\displaystyle \: { \tan}^{ - 1} \: \frac{c - a}{1 + ca} = { \tan}^{ - 1} c \: - { \tan}^{ - 1} a \:tan

−1

1+ca

c−a

=tan

−1

c−tan

−1

a

Hence

\displaystyle \: { \tan}^{ - 1} \: \frac{a - b}{1 + ab} + { \tan}^{ - 1} \: \frac{b - c}{1 + bc} + { \tan}^{ - 1} \: \frac{c - a}{1 + ca}tan

−1

1+ab

a−b

+tan

−1

1+bc

b−c

+tan

−1

1+ca

c−a

= { \tan}^{ - 1} a \: - { \tan}^{ - 1} b \: + { \tan}^{ - 1} b \: - { \tan}^{ - 1} c \: + { \tan}^{ - 1} c \: - { \tan}^{ - 1} a \:=tan

−1

a−tan

−1

b+tan

−1

b−tan

−1

c+tan

−1

c−tan

−1

a

= 0=0

Hence proved