Question

Prove that :

tan-1(1/4) + tan-1(2/9) = 1/2 Cos-1 (3/5)

Answer 1

Mark this answer as brainliest answer

Answer 2

$\bf \underline{ To \: Prove - }$


$\: \bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {tan}^{ - 1} \frac{1}{4} + {tan}^{ - 1} \frac{2}{9} = \frac{1}{2} {cos}^{ - 1} \frac{3}{5}$

$\bf \: \huge \underline{Used \: Formulas-}$

$\boxed{ \bf \: 1. \: \: {tan}^{ - 1}x = {tan}^{ - 1} (\frac{x + y}{1 - xy}) \: } \\ \boxed{ \bf \: 2. \: \: {sec}^{2} \theta \: = 1 + {tan}^{2} \theta } \\ \boxed{ \bf \: 3. \: \: cos2 \theta \: = {2 \: cos}^{2} \theta \: - 1}$

$\bf \: LHS = {tan}^{1} \frac{1}{4} + {tan}^{ - 1} \frac{2}{9} \\ \\ \\ \implies \bf \:{ {tan}^{ - 1} \: \: \: \: \: \frac{ \frac{1}{4 } + \frac{2}{9} }{1 - \frac{1}{4} \times \frac{2}{9} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge{ \{ \tiny \: {tan {}^{ - 1}x + {tan}^{ - 1}y \: = tan }^{ - 1} \: \frac{x + y}{1 - xy} } \\ \\ \\ \implies \bf \: {tan}^{ - 1} \: \: \: \: \frac{ \frac{9 + 8}{36} }{1 - \frac{1}{18} } \\ \\ \\ \implies \bf \: {tan}^{ - 1} \: \: \: \frac{ \frac{17}{36} }{ \frac{18 - 1}{18} } \\ \\ \\ \implies \bf \: {tan}^{ - 1} \: \: \frac{ \frac{17}{36} }{ \frac{17}{18} } \\ \\ \\ \implies \bf \: {tan}^{ - 1} \frac{1}{2}$

$\bf \: Let \: that \: \: \: \: \: \: \: \: \: \: \: {tan}^{1} \frac{1}{2} = \theta \: \\ \\ \: \bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{2} = tan \: \theta$

$\bf \: Now, \: we \: can \: say \: that \: \\ \\ \\ \implies {sec}^{2} \theta \: = \frac{1}{2} + \frac{1}{4} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge \{ \tiny \bf \: {sec {}^{2} \theta \: = 1 + \: tan }^{2} \theta$

$\implies \: \: {sec}^{2} \theta \: = \frac{5}{4} \\ \\ \\ \implies \bf \: \frac{1}{ {cos}^{2} \theta} = \frac{ 1 }{ \frac{5}{4} } \\ \\ \\ \bf \: \implies \: {cos}^{2} \theta \: = \frac{4}{5}$

$We \: can \: say \: that \: \\ \\ \\ cos2 \: \theta = {2} \times ( \frac{4}{5} ) \: - 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge \{ \bf \tiny cos2 \: \theta = {2cos}^{2} \theta \: - 1 \: \\ \\ \\ \implies \: \bf \: cos2 \: \theta \: = \frac{6}{5} - 1 \\ \\ \\ \implies \bf \: cos2 \: \theta \: = \frac{3}{5} \\ \\ \\ \implies \: \bf \: 2 \theta \: = {cos}^{ - 1} \frac{3}{5} \\ \\ \\ \implies \: \bf \: \theta \: = \frac{1}{2} {cos}^{ - 1} \frac{3}{5} \\ \\ \\ \bf \: now \: put \: the \: value \: of \: \theta \\ \\ \\ \boxed{ \: { \bf \: {tan}^{ - 1} \frac{1}{2} = \frac{1}{2} {cos}^{ - 1} \frac{3}{5} }}$