Question

SOLVE:-
$\tan ^{ - 1} 2x + \tan ^{ - 1} 3x = \frac{x}{4} \ \textless \ br /\ \textgreater$
Unnecessary answers will be reported...​

Answer 1

Correct Question :-

• if tan⁻¹2x + tan⁻¹3x = π/4 .. Solve for x ?

Formula used :-

• tan⁻¹A + tan⁻¹B = tan⁻¹ [ (A + B) / ( 1 - AB) ]
• tan(π/4) = 1.

Solution :-

→ tan⁻¹2x + tan⁻¹3x = π/4

Using above formula in LHS,

→ tan⁻¹ [ (2x + 3x) / ( 1 - 2x*3x) ] = π/4

→ [ 5x / (1 - 6x²) ] = tan(π/4) = 1

→ 1 - 6x² = 5x

→ 6x² + 5x - 1 = 0

Splitting The Middle Term now,

→ 6x² + 6x - x - 1 = 0

→ 6x(x +1) - 1(x + 1) = 0

→ (x + 1)(6x - 1) = 0

Putting both Equal to Zero now,

→ x + 1 = 0

→ x = (-1) .

Or,

→ (6x - 1) = 0

→ 6x = 1

→ x = (1/6) .

Hence, value of x is (-1) & (1/6).

Answer 2

${ \bold{ \boxed{ \boxed{ \mathtt{ \red{\bigstar \: ANSWER \: \bigstar}}}}}}$

${ \bold{ \underline{Given \: function} : - }} \\ \\ { \bold{ \orange{ : \implies \: {tan}^{ - 1} (2x) + {tan}^{ - 1}(3x) = \frac{\pi}{4} }}} \\ \\ { \bold{ \orange{ : \implies \: {tan}^{ - 1} ( \frac{2x + 3x}{1 - (2x)(3x)}) = \frac{\pi}{4} }}} \\ \\ { \bold{ \orange{ : \implies \: {tan}^{ - 1} ( \frac{5x}{1 - 6 {x}^{2} }) = \frac{\pi}{4} }}} \\ \\ { \bold{ \orange{ : \implies \: \frac{5x}{1 - 6 {x}^{2} } = \tan( \frac{\pi}{4} ) }}} \\ \\ { \bold{ \orange{ : \implies \: \frac{5x}{1 - 6 {x}^{2} } = 1 }}} \\ \\ { \bold{ \orange{ : \implies \: 1 - 6 {x}^{2} = 5x }}} \\ \\ { \bold{ \orange{ : \implies \: 6 {x}^{2} + 5x - 1 = 0 }}} \\ \\ { \bold{ \orange{ : \implies \: 6 {x}^{2} + 6x - x - 1 = 0}}} \\ \\ { \bold{ \orange{ : \implies \: 6x(x + 1) - 1(x + 1) = 0}}} \\ \\ { \bold{ \orange{ : \implies \: (6x - 1)(x + 1) = 0}}} \\ \\ { \bold{ \orange{ : \implies \boxed{ \: x = - 1 \:,\: x = \frac{1}{6} }}}}$

${ \bold{ \mathtt{ \green{\underline{ {used \: \: formula} } : - }}}} \\ \\ { \bold{ \pink{(1) \: {tan}^{ - 1}(x) + {tan}^{ - 1} (y) = {tan}^{ - 1} ( \frac{x + y}{1 - xy}) }}} \\ \\ { \bold{ \pink{(2) \: \tan( \frac{\pi}{4} ) = 1 }}}$