MyQuestionIcon

3

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Substitution Method to Remove Indeterminate Form

Solve the fol...

Question

Solve the following equation
${tan}^{- 1} \frac{x + 7}{x - 1} + {tan}^{- 1} \frac{x - 1}{x} = π - {tan}^{- 1} 7$ .

Open in App

Solution

${tan}^{- 1} \frac{x + 7}{x - 1} + {tan}^{- 1} \frac{x - 1}{x} = π - {tan}^{- 1} 7$ .
$\Rightarrow {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{\frac{x + 7}{x - 1} + \frac{x - 1}{x}}{1 - (\frac{x + 7}{x - 1}) (\frac{x - 1}{x})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = π - {tan}^{- 1} 7$
$\Rightarrow {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{\frac{x (x + 7) + {(x - 1)}^{2}}{x (x - 1)}}{\frac{x (x - 1) - (x + 7) (x - 1)}{x (x - 1)}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = π - {tan}^{- 1} 7$
$\Rightarrow {tan}^{- 1} [\frac{x^{2} + 7 x + x^{2} - 2 x + 1}{x^{2} - x - x^{2} - 7 x + x + 7}] = π - {tan}^{- 1} 7$
$\Rightarrow {tan}^{- 1} (\frac{2 x^{2} + 5 x + 1}{- 7 x + 7}) + {tan}^{- 1} 7 = π$
$\Rightarrow {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{\frac{{2 x}^{2} + 5 x + 1}{- 7 x + 7} + 7}{1 - (\frac{{2 x}^{2} + 5 x + 1}{- 7 x + 7}) t i m e s 7} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = π$
$\Rightarrow {tan}^{- 1} [\frac{2 x^{2} + 5 x + 1 - 49 x + 49}{- 7 x + 7 - 14 x^{2} - 35 x - 7}] = π$
$\Rightarrow \frac{2 x^{2} - 44 x + 50}{- 14 x^{2} - 42 x} tan π$
$\Rightarrow 2 x^{2} - 44 x + 50 = 0$
$\Rightarrow x^{2} - 22 x + 25 = 0$
$\Rightarrow x = \frac{22 \pm \sqrt{484 - 4 \times 25}}{2 \times 1}$
$\Rightarrow x = \frac{22 \pm 8 \sqrt{6}}{2}$
$\Rightarrow x = 11 \pm 4 \sqrt{6}$

Suggest Corrections

thumbs-up

0

Similar questions

{tan}^{- 1} (\frac{x - 1}{x - 2}) + {tan}^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4}

Q. Solve for x:

{tan}^{- 1} (x - 1) + {tan}^{- 1} x + {tan}^{- 1} (x + 1) = {tan}^{- 1} 3 x .

Solve for x the following :

{t a n}^{- 1} \frac{x - 1}{x - 2} + {t a n}^{- 1} \frac{x + 1}{x + 2} = \frac{π}{4}

{t a n}^{- 1} \frac{x - 1}{x + 1} + {t a n}^{- 1} \frac{2 x - 1}{2 x + 1} = {t a n}^{- 1} \frac{23}{36}

x

:

{tan}^{- 1} (x + 1) + {tan}^{- 1} (x - 1) = {tan}^{- 1} \frac{8}{31}

Q. Solve the following equations for x:
(i)

(ii)
(iii) tan⁻¹(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x
(iv) tan⁻¹

(\frac{1 - x}{1 + x}) - \frac{1}{2}

tan⁻¹x = 0, where x > 0
(v) cot⁻¹x − cot⁻¹(x + 2) =

\frac{π}{12}

, x > 0
(vi) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹

(\frac{8}{79})

\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Substitution Method to Remove Indeterminate Form

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Substitution Method to Remove Indeterminate Form

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon