Prove the Following Result `Tan(Cos^-1 4/5+Tan^-1 2/3)=17/6` - Mathematics

Question

Prove the following result

$\tan (\cos^{- 1} \frac{4}{5} + \tan^{- 1} \frac{2}{3}) = \frac{17}{6}$

Solution

LHS= $\tan (\cos^{- 1} \frac{4}{5} + \tan^{- 1} \frac{2}{3}) = \tan (\tan^{- 1} \frac{\sqrt{1 - {(\frac{4}{5})}^{2}}}{\frac{4}{5}} + \tan^{- 1} \frac{2}{3})$ $[∴ \cos^{- 1} x = \tan^{- 1} (\frac{\sqrt{1 - x^{2}}}{x})]$

$= \tan (\tan^{- 1} \frac{3}{4} + \tan^{- 1} \frac{2}{3})$

$= \tan [\tan^{- 1} (\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}})]$ $[∴ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})]$

$= \tan [\tan^{- 1} (\frac{\frac{17}{12}}{\frac{6}{12}})$

$= \tan [\tan^{- 1} \frac{17}{6}]$

$= \frac{17}{6} =$ RHS

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Inverse Trigonometric Functions (Simplification and Examples)

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Q 2.1 Q 1.9 Q 2.2

Chapter 4: Inverse Trigonometric Functions - Exercise 4.08 [Page 54]

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RD Sharma Mathematics [English] Class 12

Chapter 4 Inverse Trigonometric Functions
Exercise 4.08 | Q 2.1 | Page 54

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Prove the Following Result $Tan (\frac{{Cos}^{- 1} 4}{5} + \frac{{Tan}^{- 1} 2}{3}) = \frac{17}{6}$ - Mathematics

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Prove the Following Result $Tan (\frac{{Cos}^{- 1} 4}{5} + \frac{{Tan}^{- 1} 2}{3}) = \frac{17}{6}$ - Mathematics

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