Question

PLEASE GIVE EXPLAIN WITH STEPS.
The answer is option 4 but why not option 1???​

Answer 1

AnswEr :

Given Expression,

$\sf \: tan {}^{ - 1} (1) + tan {}^{ - 2} (2) + {tan}^{ - 1} (3)$

We know that,

$\sf \: tan {}^{ - 1} x + {tan}^{ - 1} (y) = tan {}^{ - 1} \bigg( \dfrac{x + y}{1 - xy} \bigg)$

Therefore,

$\sf \: tan {}^{ - 1} (2) + {tan}^{ - 1} (3) = tan {}^{ - 1} \bigg( \dfrac{2 + 3}{1 - (2)(3)} \bigg) \\ \\ \implies \sf \: \: tan {}^{ - 1} (2) + {tan}^{ - 1} (3) = tan {}^{ - 1} \bigg( - \dfrac{5}{5} \bigg) \\ \\ \implies \: \sf \: tan {}^{ - 1} (2) + {tan}^{ - 1} (3) = {tan}^{ - 1}( - 1)$

The expression can be re-written as :

$\sf \: tan {}^{ - 1} (1) + {tan}^{ - 1} ( - 1)$

Similarly,

$\longrightarrow \: \sf \: {tan}^{ - 1} \bigg( \dfrac{1 + ( - 1)}{1 - 1x - 1)} \bigg) \\ \\ \longrightarrow \sf \: {tan}^{ - 1} (0) \\ \\ \longrightarrow \sf \: tan {}^{ - 1} (tan \: \pi) \\ \\ \longrightarrow \: \pi$

Thus,

$\sf \: tan {}^{ - 1} (1) + tan {}^{ - 2} (2) + {tan}^{ - 1} (3) = \pi$

Option (4) is correct.

Note :

• The three terms tan^(-1) [1], tan^(-1) [2] and tan^(-1) [3] are positive real numbers, so their sum should also be positive. Thus,zero can't be the answer.

Answer 2

Answer:

OPTION (4) ÍŠ ÇORRECT...

HOPE IT WOULD HELP YOU.....