Inverse Circular Functions

Inverse Circular Functions


This post takes more items on Inverse Circular Functions – Chapter 18, Plane Trigonometry, SL Loney.
#12. Prove that:\[
\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{3}{5} {}-{} \tan^{-1}\frac{8}{19} = \frac{\pi}{4}
\]
Solution: \[
LHS = \tan^{-1}\left( \frac{\frac{3}{4} + \frac{3}{5}}{1 – \frac{3}{4} \times \frac{3}{5}} \right) – \tan^{-1}\frac{8}{19}
\]\[
= \tan^{-1}\frac{27}{11} {}-{} \tan^{-1}\frac{8}{19}
\]\[
= \tan^{-1}\left( \frac{\frac{27}{11} + \frac{8}{19}}{1 – \frac{27}{11} \times \frac{8}{19}} \right)
\]\[
= \tan^{-1}1 = \frac{\pi}{4} = RHS
\]

#29. Solve: \[
\tan^{-1}2x + \tan^{-1}3x = \frac{\pi}{4}
\]
Solution: Here, \[
\tan^{-1}\frac{2x + 3x}{1 – 6x^2} = \frac{\pi}{4}
\]\[
\Rightarrow \frac{5x}{1 – 6x^2} = \tan \frac{\pi}{4} = 1
\]\[
\Rightarrow 6x^2 +5x – 1 = 0
\]\[
\text or, (6x – 1)(x + 1) = 0
\]\[
\therefore x = \frac{1}{6}, – 1
\]
Now, \(- 1\) is inadmissible. Hence, \[
x = \frac{1}{6}
\]

#30. Solve: \[
\tan^{-1}\mkern-5mu\left( \frac{x – 1}{x – 2} \right) + \tan^{-1}\mkern-5mu\left( \frac{x + 1}{x + 2} \right) = \frac{\pi}{4}
\]\[
\Rightarrow \tan^{-1}\mkern-5mu\left( \frac{\frac{x – 1}{x – 2} + \frac{x + 1}{x + 2}}{1 – \frac{x^2 – 1}{x^2 – 2^2}} \right) = \frac{\pi}{4}
\]\[
\Rightarrow \frac{\frac{x – 1}{x – 2} + \frac{x + 1}{x + 2}}{1 – \frac{x^2 – 1}{x^2 – 4}} = \tan \frac{\pi}{4}
\]\[
\Rightarrow \frac{\frac{(x – 1)(x + 2) + (x + 1)(x – 2)}{x^2 – 4}}{\frac{x^2 – 4 – x^2 + 1}{x^2 – 4}} = 1
\]\[
\Rightarrow \frac{x^2 – x + 2x – 2 + x^2 + x – 2x – 2}{-3} = 1
\]\[
\Rightarrow \frac{x^2 – 4}{-3} = 1
\]\[
\Rightarrow 2x^2 = 1
\]\[
\therefore x = \pm \frac{1}{\sqrt{2}}
\]

Study: Inverse Trigonometric Functions

#32. Solve: \[
\tan^{-1}(x + 1) + \tan^{-1}(x – 1) = \tan^{-1}\frac{8}{31}
\]\[
\Rightarrow \tan^{-1}\mkern-5mu\left(\frac{x + 1 + x – 1}{1 – (x + 1)(x – 1)} \right) = \tan^{-1}\frac{8}{31}
\]\[
\Rightarrow \frac{x + 1 + x – 1}{1 – (x + 1)(x – 1)} = \frac{8}{31}
\]\[
\Rightarrow \frac{2x}{1 – (x^2 – 1)} = \frac{8}{31}
\]\[
\Rightarrow \frac{x}{2 – x^2} = \frac{4}{31}
\]\[
\Rightarrow 4x^2 + 31x – 8 = 0
\]\[
\Rightarrow (4x – 1)(x + 8) = 0
\]\[
\therefore x = \frac{1}{8}, -8
\]
Now, \(-8\) is inadmissible, hence \[
x = \frac{1}{8}
\]

#34. To solve: \[
\tan^{-1}x + 2\cot^{-1}x = \frac{2}{3}\pi
\]\[
\Rightarrow \tan^{-1}x + 2\left( \frac{\pi}{2} – \tan^{-1}x \right) = \frac{2}{3}\pi
\]\[
\Rightarrow \pi {}-{} \tan^{-1}x = \frac{2}{3}\pi
\]\[
\Rightarrow \tan^{-1}x = \frac{\pi}{3}
\]\[
\Rightarrow x = \tan \frac{\pi}{3} = \sqrt{3}
\]

#35. Solve: \[
\tan\mkern-5mu\left( \cos^{-1}x \right) = \sin\mkern-5mu\left( \cot^{-1}\frac{1}{2} \right)
\]Solution: Let \(\cos^{-1}x = \theta \Rightarrow \cos(\theta) = x \Rightarrow \tan(\theta) = \frac{\sqrt{1 – x^2}}{x}\)

Similarly, let \(\cot^{-1}\frac{1}{2} = \beta \Rightarrow \cot(\beta) = \frac{1}{2} \Rightarrow \sin(\beta) = \frac{2}{\sqrt{5}}\)

So, the eqn becomes: \[
\tan(\theta) = \sin(\beta)
\]\[
\Rightarrow \frac{\sqrt{1 – x^2}}{x} = \frac{2}{\sqrt{5}}
\]\[
\Rightarrow \frac{1 – x^2}{x^2} = \frac{4}{5}
\]\[
\Rightarrow 5 – 5x^2 = 4x^2
\]\[
\Rightarrow 9x^2 = 5
\]\[
\Rightarrow x = \pm \frac{1}{3}\sqrt{5}
\]

#41. Solve: \[
\tan^{-1}\frac{a}{x} + \tan^{-1}\frac{b}{x} + \tan^{-1}\frac{c}{x} + \tan^{-1}\frac{d}{x} = \frac{\pi}{2}
\]\[
\Rightarrow \tan^{-1}\left( \frac{\frac{a}{x} + \frac{b}{x}}{1 – \frac{a}{x} \times \frac{b}{x}} \right) + \tan^{-1}\left( \frac{\frac{c}{x} + \frac{d}{x}}{1 – \frac{c}{x} \times \frac{d}{x}} \right) = \frac{\pi}{2}
\]\[
\Rightarrow \tan^{-1}\frac{(a + b)x}{x^2 – ab} + \tan^{-1}\frac{(c + d)x}{x^2 – cd} = \frac{\pi}{2}
\]\[
\Rightarrow \tan^{-1}\frac{(a + b)x}{x^2 – ab} = \frac{\pi}{2} {}-{} \tan^{-1}\frac{(c + d)x}{x^2 – cd}
\]\[
\Rightarrow \tan^{-1}\frac{(a + b)x}{x^2 – ab} = \cot^{-1}\frac{(c + d)x}{x^2 – cd}
\]\[
\Rightarrow \frac{(a + b)x}{x^2 – ab} = \tan\mkern-5mu\left( \cot^{-1}\frac{(c + d)x}{x^2 – cd} \right)
\]\[
\Rightarrow \frac{(a + b)x}{x^2 – ab} = \tan^{-1}\mkern-5mu\left(\tan^{-1}\frac{x^2 – cd}{(c + d)x} \right)
\]\[
\Rightarrow \frac{(a + b)x}{x^2 – ab} = \frac{x^2 – cd}{(c + d)x}
\]\[
\Rightarrow x^4 – x^2(ab + ac + bc + bd + ad + cd) + abcd = 0
\]

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