Mathematics M.C.Q (1 Marks)

==> ${\cos ^{ - 1}}x = \frac{\pi }{2} - {\sin ^{ - 1}}x = \frac{\pi }{2} - \frac{\pi }{5} = \frac{{3\pi }}{{10}}$ .

$ = {\sin ^{ - 1}}\sin \left( {\frac{\pi }{3}} \right) = \frac{\pi }{3}$ .

$ = \tan \left[ {{{\tan }^{ - 1}}\frac{5}{{12}} - {{\tan }^{ - 1}}(1)} \right] = \tan {\tan ^{ - 1}}\left( {\frac{{\frac{5}{{12}} - 1}}{{1 + \frac{5}{{12}}}}} \right) = - \frac{7}{{17}}$.

$ = \tan \,\left[ {{{\tan }^{ - 1}}\frac{{\sqrt {\left( {1 - \frac{{16}}{{25}}} \right)} }}{{\frac{4}{5}}} + {{\tan }^{ - 1}}\frac{2}{3}} \right]$

$ = \tan \,\left[ {{{\tan }^{ - 1}}\left( {\frac{{\frac{3}{4} + \frac{2}{3}}}{{1 - \frac{3}{4}.\frac{2}{3}}}} \right)} \right] = \tan \,.\,{\tan ^{ - 1}}\frac{{17}}{6} = \frac{{17}}{6}$.

$ = - \sin \,\left( {{{\cos }^{ - 1}}\frac{1}{5}} \right) = - \sin \,\left( {{{\sin }^{ - 1}}\sqrt {\frac{{24}}{{25}}} } \right) = - \frac{{2\sqrt 6 }}{5}$.

$[$Since ${s^2}abc = (a + b + c)]$

Trick : Since it is an identity, so it will be true for any value of $a, b, c$.

Let $a = b = c = 1$, then $\theta = {\tan ^{ - 1}}\sqrt 3 $ $ + {\tan ^{ - 1}}\sqrt 3 $ $ + {\tan ^{ - 1}}\sqrt 3 = \pi $ $ \Rightarrow \tan \theta = 0$.

Now, $\tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{\sqrt {1 - {{(1/x)}^2}} }}{{1/x}} = \sqrt {{x^2} - 1} $.

$ \Rightarrow \,\,\cos x = \sqrt {1 - \frac{{25}}{{169}}}  = \frac{{12}}{{13}}$

==>$\,\,\cos \,\left( {{{\sin }^{ - 1}}\frac{5}{{13}}} \right) = \cos \,\,\left( {{{\cos }^{ - 1}}\frac{{12}}{{13}}} \right) = \frac{{12}}{{13}}$

$ \Rightarrow \,\,{\tan ^{ - 1}}x = \frac{\pi }{6}$

$ \Rightarrow \,\,x = \tan {30^o} = \frac{1}{{\sqrt 3 }}$.

(Putting $x = \tan \theta )$

$ = {\tan ^{ - 1}}\left[ {\frac{{\sec \theta - 1}}{{\tan \theta }}} \right] = {\tan ^{ - 1}}\left[ {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right]$

$ = {\tan ^{ - 1}}\left[ {\frac{{2\,{{\sin }^2}\frac{\theta }{2}}}{{2\,\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right]$

$ = {\tan ^{ - 1}} (\tan \frac{\theta }{2} )= \frac{\theta }{2} = \frac{1}{2}{\tan ^{ - 1}}x$.

(Putting $x = \tan \theta )$

$ = \tan \,({\sec ^{ - 1}}\,\sec \theta ) = \tan \theta = x$.

$ = {\tan ^{ - 1}}\left[ {\frac{{2\,\sin \,(\pi /4 - x/2)\,\cos \,(\pi /4 - x/2)}}{{2\,{{\cos }^2}\,(\pi /4 - x/2)}}} \right]$

$ = {\tan ^{ - 1}}\tan \,\left( {\frac{\pi }{4} - \frac{x}{2}} \right) = \frac{\pi }{4} - \frac{x}{2}$.

(Putting $x = {\rm{cos}}{\rm{ec}}\,\,\theta )$

$ = {\tan ^{ - 1}}\frac{1}{{\cot \theta }} = \theta = {\rm{cose}}{{\rm{c}}^{ - 1}}x$.

and ${\cot ^{ - 1}}3 = \beta \Rightarrow \cot \beta = 3$

${\sec ^2}({\tan ^{ - 1}}2) + {\rm{cose}}{{\rm{c}}^{\rm{2}}}({\cot ^{ - 1}}3)$

$= {\sec ^2}\alpha + {\rm{cose}}{{\rm{c}}^{\rm{2}}}\alpha $

$=1 + {\tan ^2}\alpha + 1 + {\cot ^2}\alpha $

$= 2 + {(2)^2} + {(3)^2} = 15$.

$\therefore \,\,{\sin ^{ - 1}}\frac{1}{5} = \frac{\pi }{2} - {\cos ^{ - 1}}x = {\sin ^{ - 1}}x$

$\therefore \,\,\,x = \frac{1}{5}$.

$\therefore \,\,1 + xy = 1 + \sin \,(\theta + \beta )\sin \,(\theta - \beta )$

$ = 1 + {\sin ^2}\theta - {\sin ^2}\beta = {\sin ^2}\theta + {\cos ^2}\beta $.

$ = {\sin ^{ - 1}}\left[ {\frac{1}{3}\sqrt {1 - \frac{4}{9}} + \frac{2}{3}\sqrt {1 - \frac{1}{9}} } \right] = {\sin ^{ - 1}}\,\left[ {\frac{{\sqrt 5 + 4\sqrt 2 }}{9}} \right]$

Therefore $x = \frac{{\sqrt 5 + 4\sqrt 2 }}{9}$.

==> $\theta = {\sin ^{ - 1}}[ - (\sin {240^o})]$$ = {\sin ^{ - 1}}[ - \sin ({180^o} + {60^o})]$

==> $\theta = {\sin ^{ - 1}}\sin {60^o}$ $ = {\sin ^{ - 1}}\left[ {\sin \left( {\frac{\pi }{3}} \right)} \right] = \frac{\pi }{3}$.

$ = \cos \,{\cos ^{ - 1}}\sqrt {1 - \frac{3}{4}} = \frac{1}{2}$.

$= {\tan ^{ - 1}}\left[ {\frac{{\sqrt {\left( {1 - \frac{{16}}{{25}}} \right)} }}{{\frac{4}{5}}}} \right] + {\tan ^{ - 1}}\frac{3}{5}$

$\left[ {{\rm{Since}}\,\,{{\cos }^{ - 1}}x = {{\tan }^{ - 1}}\frac{{\sqrt {(1 - {x^2})} }}{x}} \right]$

$ = {\tan ^{ - 1}}\frac{3}{4} + {\tan ^{ - 1}}\frac{3}{5} $

$= {\tan ^{ - 1}}\,\left( {\frac{{\frac{3}{4} + \frac{3}{5}}}{{1 - \frac{3}{4}.\frac{3}{5}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{27}}{{11}}} \right)$.

$= \{ {\sin ^{ - 1}}(x) + {\cos ^{ - 1}}(x)\} + \left\{ {{{\sin }^{ - 1}}\left( {\frac{1}{x}} \right) + {{\cos }^{ - 1}}\left( {\frac{1}{x}} \right)} \right\}$

$ = \frac{\pi }{2} + \frac{\pi }{2} = \pi $.

$ = {\tan ^{ - 1}}1 + \pi + {\tan ^{ - 1}}\left( {\frac{5}{{ - 5}}} \right)$

$ = {\tan ^{ - 1}}1 + \pi - {\tan ^{ - 1}}1 = \pi $.

and $\sin \theta = \frac{1}{{\sqrt {1 + {{\cot }^2}\theta } }} = \frac{1}{{\sqrt {1 + (9/16)} }} = \frac{4}{5}$

Hence ${\cot ^{ - 1}}\frac{3}{4} + {\sin ^{ - 1}}\frac{5}{{13}} = {\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{5}{{13}}$

$ = {\sin ^{ - 1}}\left[ {\frac{4}{5}.\sqrt {1 - \frac{{25}}{{169}}} + \frac{5}{{13}}.\,\sqrt {1 - \frac{{16}}{{25}}} } \right]$

$ = {\sin ^{ - 1}}\left[ {\frac{4}{5}.\frac{{12}}{{13}} + \frac{5}{{13}}.\frac{3}{5}} \right]$

$ = {\sin ^{ - 1}}\left[ {\frac{{48 + 15}}{{65}}} \right] = {\sin ^{ - 1}}\frac{{63}}{{65}}$.

$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left( {\frac{{x - y}}{{1 + xy}}} \right) = {\tan ^{ - 1}}A$.

Hence $A = \frac{{x - y}}{{1 + xy}}$.

$ = {\tan ^{ - 1}}(a) - {\tan ^{ - 1}}(b) + {\tan ^{ - 1}}(b) - {\tan ^{ - 1}}(c)$

$ = {\tan ^{ - 1}}(a) - {\tan ^{ - 1}}(c)$.

$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left( {\frac{{2x + 3x}}{{1 - (2x)\,(3x)}}} \right) = \frac{\pi }{4}\,$

$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left( {\frac{{5x}}{{1 - 6{x^2}}}} \right)\, = {\tan ^{ - 1}}(1)$

$ \Rightarrow \,\,\frac{{5x}}{{1 - 6{x^2}}} = 1\,$

$ \Rightarrow \,\,1 - 6{x^2} = 5x$

$\, \Rightarrow \,\,6{x^2} + 5x - 1 = 0$

$ \Rightarrow \,\,(x + 1)\,\left( {x - \frac{1}{6}} \right) = 0\,$

$ \Rightarrow \,\,x = - 1,\,\,\frac{1}{6}$

But $-1$ does not hold.

Trick : Check with the options. 

Obviously the equation holds for $x = \frac{1}{6}$, but not for $-1$.

$ \Rightarrow \,\,{\cot ^{ - 1}}x + {\tan ^{ - 1}}3 = \frac{\pi }{2}\,\, $

$\Rightarrow \,\,{\tan ^{ - 1}}\frac{1}{x} + {\tan ^{ - 1}}3 = \frac{\pi }{2}$

$ \Rightarrow \,\,{\tan ^{ - 1}}\left( {\frac{{\frac{1}{x} + 3}}{{1 - \frac{1}{x}.3}}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{0}} \right)$

$ \Rightarrow \,\,\frac{{3x + 1}}{{x - 3}} = \frac{1}{0}\,\, \Rightarrow \,\,x = 3$

Aliter : As we know that, ${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2},$

therefore for the given question, $ x$ should be $3.$

$ = {\sin ^{ - 1}}\frac{{24}}{{25}} + {\cos ^{ - 1}}\frac{{24}}{{25}} = \frac{\pi }{2}$.

$ = \cos \,\{ {\tan ^{ - 1}}(1)\} = \cos \frac{\pi }{4} = \frac{1}{{\sqrt 2 }}$.

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

$ = {\cot ^{ - 1}}y - {\cot ^{ - 1}}x + {\cot ^{ - 1}}z - {\cot ^{ - 1}}y$$ + {\cot ^{ - 1}}x - {\cot ^{ - 1}}z = 0$.

Note: Students should remember this question as a formula.

$ \Rightarrow {\cos ^{ - 1}}\frac{3}{5} - {\cos ^{ - 1}}\sqrt {1 - \frac{{16}}{{25}}} = {\cos ^{ - 1}}x$

$ \Rightarrow {\cos ^{ - 1}}\frac{3}{5} - {\cos ^{ - 1}}\frac{3}{5} = {\cos ^{ - 1}}x$

$ \Rightarrow {\cos ^{ - 1}}x = 0 \Rightarrow x = 1$.

$ = {\cot ^{ - 1}}\left( {\frac{{3 \times 2 - 1}}{{3 + 2}}} \right) = {\cot ^{ - 1}}(1) = \frac{\pi }{4}$.

${\tan ^{ - 1}}\frac{{1 - {x^2}}}{{2x}} + {\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}}$

$ = {\tan ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{2\tan \theta }}} \right) + {\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$ = {\tan ^{ - 1}}(\cot 2\theta ) + {\cos ^{ - 1}}(\cos 2\theta )$

$ = \frac{\pi }{2} - 2\theta + 2\theta = \frac{\pi }{2}$.

$ \Rightarrow \frac{\pi }{2} - {\sin ^{ - 1}}x + \frac{\pi }{2} - {\sin ^{ - 1}}y = 2\pi $

$ \Rightarrow \pi - ({\sin ^{ - 1}}x + {\sin ^{ - 1}}y) = 2\pi $

$ \Rightarrow {\sin ^{ - 1}}x + {\sin ^{ - 1}}y = - \pi $

$ = {\cot ^{ - 1}}(2) + {\cot ^{ - 1}}(3) = {\cot ^{ - 1}}\left( {\frac{{2 \times 3 - 1}}{{3 + 2}}} \right) = {\cot ^{ - 1}}(1) = \frac{\pi }{4}$.

$= \frac{\pi }{3} + \frac{{2\pi }}{6} = \frac{{2\pi }}{3}$.

$ = {\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}}\left( {\frac{{425}}{{425}}} \right) = {\tan ^{ - 1}}(1) = \frac{\pi }{4}$.

$ \Rightarrow \frac{\pi }{2} - {\cos ^{ - 1}}x + \frac{\pi }{2} - {\cos ^{ - 1}}y = \frac{{2\pi }}{3}$

==> ${\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \pi - \frac{{2\pi }}{3} = \frac{\pi }{3}$.

$={\tan ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{1}{2}.2{\tan ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{1}{2}{\tan ^{ - 1}}\frac{{2(1/2)}}{{1 - (1/4)}}$

$ = \frac{1}{2}{\tan ^{ - 1}}\frac{4}{3} = \frac{1}{2}{\cos ^{ - 1}}\frac{3}{5}$.

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

$ = {\tan ^{ - 1}}\left( {\frac{{25}}{{25}}} \right) = \frac{\pi }{4}$.

$ = {\cos ^{ - 1}}\left( {\frac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\frac{{1 - 1/25}}{{1 + 1/25}}} \right)$

$ = {\cos ^{ - 1}}\left( {\frac{{15}}{{17}}} \right) + {\cos ^{ - 1}}\left( {\frac{{12}}{{13}}} \right)$

$ = {\cos ^{ - 1}}\left( {\frac{{15}}{{17}} \times \frac{{12}}{{13}} - \sqrt {1 - {{\left( {\frac{{15}}{{17}}} \right)}^2}} \sqrt {1 - {{\left( {\frac{{12}}{{13}}} \right)}^2}} } \right)$

$ = {\cos ^{ - 1}}\left( {\frac{{140}}{{221}}} \right)$.

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

$= {\tan ^{ - 1}}\left( {\frac{{25}}{{25}}} \right) = {\tan ^{ - 1}}1 = \frac{\pi }{4}$.