Inverse Trigonometric Functions Test - 19

We have,

\(tan^{-1}\) x + \(tan^{-1}\)y = \(\frac{4\pi}{5}\),

⇒ \(\frac{\pi}{2}\) - \(cot^{-1} \)x + \(\frac{\pi}{2}\) - \(cot^{-1} \)y = \(\frac{4\pi}{5}\)

⇒ -(\(cot^{-1} \)x + \(cot^{-1} \)y) = \(\frac{4\pi}{5}\) - \(\pi\)   [\(\because\) \(tan^{-1}\)x + \(cot^{-1} \)x = \(\frac{\pi}{2}\)]

⇒ \(cot^{-1} \)x + \(cot^{-1} \)y = -\((-\frac{\pi}{5})\)

⇒ \(cot^{-1} \)x + \(cot^{-1} \)y = \(\frac{\pi}{5}\)

We have,

\(sin^{-1}\)\((\frac{2a}{1+a^2})\) + \(cos^{-1}\)\((\frac{1-a^2}{1+a^2})\) = \(tan^{-1}\)\((\frac{2x}{1-x^2})\)

Let a = tan\(\theta\) ⇒ \(\theta\) = \(tan^{-1}\)a

\(\therefore\) \(sin^{-1}\)\((\frac{2tan\theta}{1+tan^2\theta})\) + \(cos^{-1}\)\((\frac{1-tan^2\theta}{1+tan^2\theta})\) = \(tan^{-1}\)\((\frac{2x}{1-x^2})\)

⇒ \(sin^{-1}\)sin2\(\theta\) + \(cos^{-1}\)cos2\(\theta\) = \(tan^{-1}\)\(\frac{2x}{1-x^2}\)

⇒ 2\(\theta\) + 2\(\theta\) = \(tan^{-1}\)\(\frac{2x}{1-x^2}\)

⇒ 4\(tan^{-1}\)a = \(tan^{-1}\)\(\frac{2x}{1-x^2}\)

⇒ 2.2\(tan^{-1}\)a = \(tan^{-1}\)\(\frac{2x}{1-x^2}\)

⇒ 2.\(tan^{-1}\)\(\frac{2a}{1-a^2}\) = \(tan^{-1}\)\(\frac{2x}{1-x^2}\)   [\(\because\) 2\(tan^{-1}\)x = \(tan^{-1}\)\(\frac{2x}{1-x^2}\)]

⇒ \(tan^{-1}\)\(\Bigg[\frac{2(\frac{2a}{1-a^2})}{1-(\frac{2a}{1-a^2})^2}\Bigg]\)=\(tan^{-1}\)\(\frac{2x}{1-x^2}\) 

\(\therefore\) x = \(\frac{2a}{1-a^2}\)

We have,

tan(\(\frac{1}{2}\)\(cos^{-1}\)\(\frac{2}{\sqrt5}\))

⇒ Let \(\frac{1}{2}\)\(cos^{-1}\)\(\frac{2}{\sqrt5}\) = \(\theta\)

⇒ \(cos^{-1}\)\(\frac{2}{\sqrt5}\) = 2\(\theta\) ⇒ cos 2\(\theta\) = \(\frac{2}{\sqrt5}\)

\(\therefore\) (1 - 2\(sin^2\theta\)) = \(\frac{2}{\sqrt5}\)

⇒ 2\(sin^2\theta\) = 1 - \(\frac{2}{\sqrt5}\)

⇒ \(sin^2\theta\) = \(\frac{1}{2}\) - \(\frac{1}{\sqrt5}\)

⇒ sin \(\theta\) = \(\sqrt{\frac{1}{2}-\frac{1}{\sqrt5}}\)

\(\therefore\) \(cos^2\theta\) = 1 - \(sin^2\theta\)

= 1 - \(\frac{1}{2}\) + \(\frac{1}{\sqrt5}\) = \(\frac{1}{2}\) + \(\frac{1}{\sqrt5}\)

⇒ cos \(\theta\) = \(\sqrt{\frac{1}{2}+\frac{1}{\sqrt5}}\)

\(\therefore\) tan \(\theta\) = \(\sqrt{\frac{\frac{1}{2}-\frac{1}{\sqrt5}}{\frac{1}{2}+\frac{1}{\sqrt5}}}\) = \(\sqrt{\frac{\sqrt5-2}{\sqrt5+2}}\) 

\(\therefore\) tan(\(\frac{1}{2}\)\(cos^{-1}\)\(\frac{2}{\sqrt5}\)) = tan\(tan^{-1}\)\(\sqrt{\frac{\sqrt5-2}{\sqrt5+2}}\)

= \(\sqrt{\frac{\sqrt5-2}{\sqrt5+2}.\frac{\sqrt5-2}{\sqrt5-2}}\)

= \(\sqrt{\frac{(\sqrt5-2)^2}{5-4}}\) = \(\sqrt5 -2\)

We have,

2\(tan^{-1}\)x + \(sin^{-1}\)\((\frac{2x}{1+x^2})\)

Let x = tan \(\theta\)

\(\therefore\) 2\(tan^{-1}\)tan \(\theta\) + \(sin^{-1}\)\(\frac{2tan\theta}{1+tan^2\theta}\)   [\(\because\) \(tan^{-1}\)(tan x) = x]

= 2\(\theta\) + \(sin^{-1}\)sin2\(\theta\)  [\(\because\) sin2\(\theta\) = \(\frac{2tan\theta}{1+tan^2\theta}\)]

= 2\(\theta\) + 2\(\theta\)  [\(\because\) \(sin^{-1}\)(sin x) = x]

= 4\(\theta\)  [\(\because\) \(\theta\) = \(tan^{-1}\)x]

= 4\(tan^{-1}\)x

We have

\(cos^{-1 }\)\(\alpha\) + \(cos^{-1 }\)\(\beta\) + \(cos^{-1 }\)\(\gamma\) = 3\(\pi\),

We know that, 0 \(\leq\) \(cos^{-1 }\)x \(\leq\) \(\pi\)

⇒ \(cos^{-1 }\)\(\alpha\) + \(cos^{-1 }\)\(\beta\) + \(cos^{-1 }\)\(\gamma\) = 3\(\pi\)

If and only if, \(cos^{-1 }\)\(\alpha\) = \(cos^{-1 }\)\(\beta\) = \(cos^{-1 }\)\(\gamma\) = \(\pi\)

⇒ cos \(\pi\) = \(\alpha\) = \(\beta\) = \(\gamma\)

⇒ -1 = \(\alpha\) = \(\beta\) = \(\gamma\)

⇒ \(\alpha\) = \(\beta\) = \(\gamma\) = -1

\(\therefore\) \(\alpha\)(\(\beta\) + \(\gamma\)) + \(\beta\)(\(\gamma\) + \(\alpha\)) + \(\gamma\)(\(\alpha\) + \(\beta\))

= -1(-1-1) -1(1-1)-1(-1-1)

= 2 + 2 + 2 = 6

We have,

\(\sqrt{1+cos2x}\) = \(\sqrt2\)\(cos^{-1}\) (cos x) in \([\frac{\pi}{2},\pi]\)

⇒ \(\sqrt{1+2cos^2x-1}\) = \(\sqrt2\)\(cos^{-1}\)(cos x)

⇒ \(\sqrt2\)cos = \(\sqrt2\)\(cos^{-1}\)(cos x)

⇒ cos x = \(cos^{-1}\)(cos x)

⇒ cos x = x     [\(\because\)\(cos^{-1}\)(cos x) = x]

Which is not true for any real value of x.

Hence, there is no solution possible for the given equation.

\(sin^{-1}\)sin(-30°) = \(-\frac{\pi}{6}\)

Let \(sin^{-1}\)x = \(\theta\) ⇒ sin\(\theta\) = x Now

1 + \(cot^2\theta\) = \(cosec^2\theta\) = \(\frac{1}{x^2}\)

Hence 1 + \(cot^2\)\((sin^{-1}x)\) = \(\frac{1}{x^2}\)