Prove `2 Tan^(-1) 1/2 + Tan^(-1) 1/7 = Tan^(-1) 31/17` - Mathematics

प्रश्न

Prove $2 \tan^{- 1} \frac{1}{2} + \tan^{- 1} \frac{1}{7} = \tan^{- 1} \frac{31}{17}$

उत्तर

Tp prove $2 \tan^{- 1} \frac{1}{2} + \tan^{- 1} \frac{1}{7} = \tan^{- 1} \frac{31}{17}$

L.H.S = $2 \tan^{- 1} \frac{1}{2} + \tan^{- 1} \frac{1}{7}$

= $\tan^{- 1} \frac{2 . \frac{1}{2}}{1 - {(\frac{1}{2})}^{2}} + \tan^{- 1} \frac{1}{7}$ $[2 \tan^{- 1} x = \tan^{- 1} \frac{2 x}{1 - x^{2}}]$

$= \tan^{- 1} \frac{1}{(\frac{3}{4})} + \tan^{- 1} \frac{1}{7}$

$= \tan^{- 1} \frac{4}{3} + \tan^{- 1} \frac{1}{7}$

= $\tan^{- 1} \frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} . \frac{1}{7}}$ $[\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \frac{x + y}{1 - x y}]$

$= \tan^{- 1} \frac{\frac{28 + 3}{21}}{\frac{21 - 4}{21}}$

= $\tan^{- 1} \frac{31}{17}$ = R.H.S

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Properties of Inverse Trigonometric Functions

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Q 4 Q 3 Q 5

अध्याय 2: Inverse Trigonometric Functions - Exercise 2.2 [पृष्ठ ४७]

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एनसीईआरटी Mathematics [English] Class 12

अध्याय 2 Inverse Trigonometric Functions
Exercise 2.2 | Q 4 | पृष्ठ ४७

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Prove $2 \frac{{Tan}^{- 1} 1}{2} + \frac{{Tan}^{- 1} 1}{7} = \frac{{Tan}^{- 1} 31}{17}$ - Mathematics

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