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This is the second of our many quizzes on Inverse trigonometric ratios. Click on this to access more such competitive quizzes on Trigonometry.

Inverse trigonometric ratios Quiz 2

mathematics

 Evaluate: ${{\sin }^{-1}}\left[ \sin \left( \dfrac{4\pi }{7} \right) \right]$.

 Only using the identity ${{\sin }^{-1}}\left[ \sin x \right]=x$ will not give the answer as $\dfrac{4\pi }{7}\notin \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$, first subtract the given angle from $\pi $, and then use that identity.

${{\sin }^{-1}}\left[ \sin \left( \dfrac{4\pi }{7} \right) \right]$ $ \\ $ We know that ${{\sin }^{-1}}\left[ \sin x \right]=x$ $\therefore {{\sin }^{-1}}\left[ \sin \left( \dfrac{4\pi }{7} \right) \right]=\dfrac{4\pi }{7}$, but $\dfrac{4\pi }{7}\notin \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ So, $\sin \left( \dfrac{4\pi }{7} \right)=\sin \left[ \pi -\dfrac{4\pi }{7} \right]=\sin \left( \dfrac{3\pi }{7} \right)$ Therefore, ${{\sin }^{-1}}\left[ \sin \left( \dfrac{4\pi }{7} \right) \right]={{\sin }^{-1}}\left[ \sin \left( \dfrac{3\pi }{7} \right) \right]=\dfrac{3\pi }{7}$

 Find the combined value of ${{\tan }^{-1}}\dfrac{4}{9}$ and ${{\tan }^{-1}}\dfrac{5}{8}$.

 First use the identity $\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$, then simplify the obtained expression.

  ${{\tan }^{-1}}\dfrac{4}{9}+{{\tan }^{-1}}\dfrac{5}{8} \\ ={{\tan }^{-1}}\dfrac{\dfrac{4}{9}+\dfrac{5}{8}}{1-\dfrac{4}{9}\times \dfrac{5}{8}}={{\tan }^{-1}}\dfrac{32+45}{72-20} \\ ={{\tan }^{-1}}\dfrac{77}{52} $

 Find the value of ${{\cot }^{-1}}\left( \sec x-\tan x \right)$ if $x\in \left( -\dfrac{3\pi }{2},\dfrac{\pi }{2} \right)$.

 First simplify the given expression, then use the identities $\cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}\text{ and }\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$.

${{\cot }^{-1}}\left( \sec x-\tan x \right)={{\cot }^{-1}}\left( \dfrac{1-\sin x}{\cos x} \right) \\ ={{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)={{\tan }^{-1}}\left[ \dfrac{{{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}}{{{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2}-2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right] \\  ={{\tan }^{-1}}\left[ \dfrac{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}{{{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}^{2}}} \right] \\ ={{\tan }^{-1}}\left[ \dfrac{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)} \right]={{\tan }^{-1}}\left[ \dfrac{1+\tan \dfrac{x}{2}}{1-\tan \dfrac{x}{2}} \right] \\ ={{\tan }^{-1}}\left[ \tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right) \right]=\dfrac{\pi }{4}+\dfrac{x}{2} $

 Find the difference between ${{\sin }^{-1}}\dfrac{5}{13}$ and ${{\sin }^{-1}}\dfrac{7}{25}$.

 First let ${{\sin }^{-1}}\dfrac{5}{13}=x$ and ${{\sin }^{-1}}\dfrac{7}{25}=y$, use these two for the values of cosine and tangent function, then at last use the identity $\cos \left( x-y \right)=\cos x\cos y+\sin x\sin y$.

Let ${{\sin }^{-1}}\dfrac{5}{13}=x$ and ${{\sin }^{-1}}\dfrac{7}{25}=y$ $\sin x=\dfrac{5}{13}\text{ and }\sin y=\dfrac{7}{25}$ Now, $\cos x=\sqrt{1-{{\left( \dfrac{5}{13} \right)}^{2}}}=\dfrac{12}{13}\text{ and }\cos y=\sqrt{1-{{\left( \dfrac{7}{25} \right)}^{2}}}=\dfrac{24}{25}$ $\cos \left( x-y \right)=\cos x\cos y+\sin x\sin y \\ \Rightarrow \cos \left( {{\sin }^{-1}}\dfrac{5}{13}-{{\sin }^{-1}}\dfrac{7}{25} \right)=\dfrac{12}{13}\times \dfrac{24}{25}+\dfrac{5}{13}\times \dfrac{7}{25} \\ \therefore \cos \left( {{\sin }^{-1}}\dfrac{5}{13}-{{\sin }^{-1}}\dfrac{7}{25} \right)=\dfrac{288+35}{325}=\dfrac{323}{325} $

 Find the value of ${{\sin }^{-1}}\dfrac{21}{29}+{{\cos }^{-1}}\dfrac{8}{17}+{{\tan }^{-1}}\dfrac{63}{31}.$

 First let ${{\sin }^{-1}}\dfrac{21}{29}=x$ and ${{\cos }^{-1}}\dfrac{8}{17}=y$ use these two for the values of cosine and tangent function, then at last use the identity $\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$.

Let ${{\sin }^{-1}}\dfrac{21}{29}=x\Rightarrow \sin x=\dfrac{21}{29}$ and ${{\cos }^{-1}}\dfrac{8}{17}=y\Rightarrow \cos y=\dfrac{8}{17}$ $\therefore \cos x=\sqrt{1-{{\left( \dfrac{21}{29} \right)}^{2}}}=\dfrac{20}{29}$ and $\sin y=\sqrt{1-{{\left( \dfrac{8}{17} \right)}^{2}}}=\dfrac{15}{17}$ Now, $\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}=\dfrac{\dfrac{21}{20}+\dfrac{15}{8}}{1-\dfrac{21}{20}\times \dfrac{15}{8}}=\dfrac{\dfrac{117}{40}}{-\dfrac{31}{32}}=-\dfrac{9}{155}$ $\tan (x+y)=-\tan z \\ \Rightarrow \tan (x+y)=\tan (-z)=\tan (\pi -z)\therefore x+y=\pi -z $ So, $x+y+z=\pi \Rightarrow {{\sin }^{-1}}\dfrac{21}{29}+{{\cos }^{-1}}\dfrac{8}{17}+{{\tan }^{-1}}\left(\dfrac{9}{155}\right)=\pi $