Maths M.C.Q (1 Marks)

$ \Rightarrow $ $\tan 2\theta = \tan \frac{\pi }{4}$

$\Rightarrow 2\theta = n\pi + \frac{\pi }{4}$

$\Rightarrow \theta = \frac{{n\pi }}{2} + \frac{\pi }{8}$.

$ \Rightarrow $ $\sin \left( {\theta + \frac{\pi }{3}} \right) = \frac{1}{{\sqrt 2 }} = \sin \left( {\frac{\pi }{4}} \right)$

$ \Rightarrow $ $\theta = n\pi + {( - 1)^n}\frac{\pi }{4} - \frac{\pi }{3}$.

$ \Rightarrow $ ${\cos ^2}\theta + 2\cos \theta - \frac{5}{4} = 0$

$ \Rightarrow $ $\cos \theta = \frac{{ - 2 \pm \sqrt {4 + 5} }}{2} = - 1 \pm \frac{3}{2}$

Since $|\cos \theta |\, \le 1$, 

hence $\cos \theta = - 1 - \frac{3}{2}$ is ruled out.

$ \Rightarrow $ $\cos \theta = - 1 + \frac{3}{2} = \frac{1}{2} = \cos \left( {\frac{\pi }{3}} \right)$

$ \Rightarrow $ $\theta = 2n\pi \pm \frac{\pi }{3}$.

$\Rightarrow \frac{{\sqrt 2 }}{{\cos \theta }} + \frac{{\sin \theta }}{{\cos \theta }} = 1$

$ \Rightarrow $ $\sin \theta - \cos \theta = - \sqrt 2 $

Dividing by $\sqrt 2 $ on both sides, we get 

$\frac{1}{{\sqrt 2 }}\sin \theta - \frac{1}{{\sqrt 2 }}\cos \theta = - 1$

$ \Rightarrow $ $\frac{1}{{\sqrt 2 }}\cos \theta - \frac{1}{{\sqrt 2 }}\sin \theta = 1 $

$\Rightarrow \cos \,\left( {\theta + \frac{\pi }{4}} \right) = \cos (0)$

$ \Rightarrow $ $\theta + \frac{\pi }{4} = 2n\pi \pm 0$

$\Rightarrow \theta = 2n\pi - \frac{\pi }{4}$.

$i.e.,$ $\sin \theta = 0 \Rightarrow \theta = n\pi $

or $\frac{1}{{\cos \theta }} = - 2 $

$\Rightarrow \cos \theta = - \frac{1}{2}$

$\Rightarrow \theta = 2n\pi \pm \left( {\frac{{2\pi }}{3}} \right)$.

$ \Rightarrow $ $\frac{{\tan 2\theta + \tan 3\theta }}{{1 - \tan 2\theta \tan 3\theta }} = \frac{1}{{\sqrt 3 }}$ 

==> $\tan 5\theta = \tan \frac{\pi }{6}$

$ \Rightarrow $ $5\theta = n\pi + \frac{\pi }{6} $

$\Rightarrow \theta = \left( {n + \frac{1}{6}} \right)\frac{\pi }{5}$.

==> $\tan 2\theta = \tan {\rm{ }}\left( {\frac{\pi }{2} - \theta } \right)$

$ \Rightarrow $ $2\theta = n\pi + \frac{\pi }{2} - \theta$

$\Rightarrow \theta = \frac{{n\pi }}{3} + \frac{\pi }{6}$.

also ${\sec ^2}\theta - {\tan ^2}\theta = 1$

$ \Rightarrow $ ${\tan ^2}\theta = \frac{1}{3} = {\tan ^2}\left( {\frac{\pi }{6}} \right) $

$\Rightarrow \theta = n\pi \pm \frac{\pi }{6}$.

==> $\sin 4\theta = 2\sin (4\theta )\sin (3\theta )$

$ \Rightarrow $ $\sin 4\theta = 0 $

$\Rightarrow $ $4\theta = n\pi $

or $\sin 3\theta = \frac{1}{2} = \sin \left( {\frac{\pi }{6}} \right)$

$ \Rightarrow $ $3\theta = n\pi + {( - 1)^n}\frac{\pi }{6} $

$\Rightarrow \theta = \frac{{n\pi }}{4},\,\frac{{n\pi }}{3} + {( - 1)^n}\frac{\pi }{{18}}$.

$\Rightarrow {\cos ^2}\theta - {\sin ^2}\theta = \frac{1}{2}$

$ \Rightarrow $ $\cos 2\theta = \frac{1}{2} = \cos \left( {\frac{\pi }{3}} \right)$

$ \Rightarrow $ $2\theta = 2n\pi \pm \frac{\pi }{3} $

$\Rightarrow \theta = n\pi \pm \frac{\pi }{6}$.

==> $\frac{2}{{\sin \theta }} = \frac{1}{{\sin \theta \cos \theta }}$

$ \Rightarrow $ $\cos \theta = \frac{1}{2}$ or $\sin \theta = 0$

$ \Rightarrow $ $\theta = 2n\pi \pm \frac{\pi }{3}$ or $\theta = n\pi $.

$ \Rightarrow $ $\tan \theta (\tan \theta - 1) - \sqrt 3 (\tan \theta - 1) = 0$

$ \Rightarrow $$(\tan \theta - \sqrt 3 )\,\,(\tan \theta - 1) = 0$

$ \Rightarrow $$\theta = n\pi + \frac{\pi }{3}$, $n\pi + \frac{\pi }{4}$.

$ \Rightarrow $ $4{\cos ^2}\theta - 2\,(\sqrt 3 + 1)\cos \theta + \sqrt 3 = 0$

$ \Rightarrow $ $\cos \theta = \frac{{2(\sqrt 3 + 1) \pm \sqrt {4{{(\sqrt 3 + 1)}^2} - 16\sqrt 3 } }}{8}$

$ \Rightarrow $ $\cos \theta = \frac{{\sqrt 3 }}{2}{\rm{ or}}\,\,{\rm{1/2}}$

$\Rightarrow \theta = 2n\pi \pm \frac{\pi }{6}$ or $2n\pi \pm \pi /3$.

$ \Rightarrow \frac{{\cos \theta }}{{\sin \theta }} + \frac{{\cos \{ (\pi /4) + \theta \} }}{{\sin \{ (\pi /4) + \theta \} }} = 2$

$ \Rightarrow $ $\sin \left( {\frac{\pi }{4} + 2\theta } \right) = 2\sin \theta \sin \left( {\frac{\pi }{4} + \theta } \right)$

$ \Rightarrow $ $\sin \left( {\frac{\pi }{4} + 2\theta } \right) + \cos \left( {\frac{\pi }{4} + 2\theta } \right) = \frac{1}{{\sqrt 2 }}$

$ \Rightarrow $ $\cos 2\theta = \frac{1}{2} $

$\Rightarrow 2\theta = 2n\pi \pm \frac{\pi }{3} $

$\Rightarrow \theta = n\pi \pm \frac{\pi }{6}$.

$\cos \theta = \frac{{ - 3 \pm \sqrt {9 + 8} }}{4} = \frac{{ - 3 \pm \sqrt {17} }}{4}$

$ \Rightarrow $ $\theta = 2n\pi \pm {\cos ^{ - 1}}\left( {\frac{{ - 3 + \sqrt {17} }}{4}} \right)$, (Taking $+ve$ sign).

==>$3\sin (A - {15^o})\cos (A + {15^o})$$ = \cos (A - {15^o})\sin (A + {15^o})$

$ \Rightarrow $ $2\sin (A - {15^o})\cos (A + {15^o}) = \frac{1}{2}$

$ \Rightarrow $ $\sin 2A - \sin {30^o} = \frac{1}{2}$

$ \Rightarrow $$2A = 2n\pi + \frac{\pi }{2}$

$ \Rightarrow $ $A = n\pi + \frac{\pi }{4}$.

$ \Rightarrow $ ${\tan ^2}\theta - 2\tan \theta + 1 = 0$

$ \Rightarrow $ $\tan \theta = 1 = \tan \frac{\pi }{4}$ 

$ \Rightarrow $ $\theta = n\pi + \frac{\pi }{4}$.

$\therefore $ ${\sin ^2}x = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2}$ 

==> ${\sin ^2}x = {\sin ^2}\pi /3$

==> $x = n\pi \pm \pi /3$.

$ \Rightarrow \tan \theta + \cot \theta = 2$

$ \Rightarrow $ $\sin 2\theta = 1 = \sin \frac{\pi }{2} $

$\Rightarrow \theta = n\pi + \frac{\pi }{4}$.

$ \Rightarrow $ $(2\sin \theta + 1)\,\,(\sin \theta - 2) = 0$

$ \Rightarrow $ $\sin \theta = - \frac{1}{2}$ , $(\therefore \,\sin \theta  \ne 2)$ 

$ \Rightarrow $ $\sin \theta = \sin \left( {\frac{{ - \pi }}{6}} \right)$

$ \Rightarrow $ $\theta = n\pi + {( - 1)^n}\left( {\frac{{ - \pi }}{6}} \right) $

$\Rightarrow \theta = n\pi + {( - 1)^{n + 1}}\frac{\pi }{6}$

$ \Rightarrow $ $\theta = n\pi + {( - 1)^n}\frac{{7\pi }}{6}$, $\left\{ \because {\,\,\frac{{ - \pi }}{6}{\rm{\,\,is \,\,equivalent \,\,to\,\, }}\frac{{7\pi }}{6}} \right\}$.

$ \Rightarrow $$\frac{{{{\sin }^2}\theta }}{{\cos \theta }} + \sqrt 3 \tan \theta = 0$

$ \Rightarrow $ $\tan \theta \sin \theta + \sqrt 3 \tan \theta = 0$

$\tan \theta (\sin \theta + \sqrt 3 ) = 0$

$ \Rightarrow $ $\tan \theta = 0$ 

$ \Rightarrow $$\theta = n\pi ,\,n \in Z$.

$ \Rightarrow 2\sin 4\theta \cos 2\theta + \sin 4\theta = 0$

$ \Rightarrow $$\sin 4\theta (2\cos 2\theta + 1) = 0$

$ \Rightarrow $ $2\cos 2\theta = - 1$ 

$ \Rightarrow $ $\cos 2\theta = - \frac{1}{2}$

$ \Rightarrow $ $2\theta = 2n\pi \pm \frac{{2\pi }}{3} $

$\Rightarrow \theta = n\pi \pm \frac{\pi }{3}$

and $\sin 4\theta = 0 \Rightarrow 4\theta = n\pi \Rightarrow \theta = \frac{{n\pi }}{4}$

$\theta = \frac{{n\pi }}{4}$ or $n\pi \pm \frac{\pi }{3}$.

$3(1 - {\cos ^2}x) + 10\cos x - 6 = 0$

On solving, $(\cos x - 3)\,(3\cos x - 1) = 0$

Either $\cos x = 3$, (which is not possible) or 

$\cos x =\frac{1}{3}$ $ \Rightarrow \,\,x = 2n\pi \pm {\cos ^{ - 1}}(1/3)$.

==> $(\cos \theta + \cos 3\theta ) + \cos 2\theta = 0$

==> $2\cos 2\theta \cos \theta + \cos 2\theta = 0$

==> $\cos 2\theta (2\cos \theta + 1) = 0$ 

==> $\cos 2\theta = 0 = \cos \frac{\pi }{2}$

==> $\theta = \frac{\pi }{4}$

==> $\theta = 2m\pi \pm \frac{\pi }{4}$

or $\cos \theta = \frac{{ - 1}}{2} = \cos \frac{{2\pi }}{3}$ 

==> $\theta = 2m\pi \pm \frac{{2\pi }}{3}$.

==> $2\sqrt 3 {\sin ^2}\theta + \sin \theta - 2\sqrt 3 = 0$

==> $\sin \theta = \frac{{ - 1 \pm 7}}{{4\sqrt 3 }}$

$\Rightarrow \sin \theta = \frac{{ - 8}}{{4\sqrt 3 }}{\rm{,}}$ (Impossible)

and $\sin \theta = \frac{6}{{4\sqrt 3 }} = \frac{{\sqrt 3 }}{2}$

==> $\theta = n\pi + {( - 1)^n}\frac{\pi }{3}$.

${\cos ^2}(A + B) + {\sin ^2}(A + B) + \cos 2B = 0$

$1 + \cos 2B = 0$ or $\cos 2B = \cos \pi $

or $2B = 2n\pi + \pi $ or $B = (2n + 1)\frac{\pi }{2},\,\,n \in Z.$

$4{\sin ^4}x = 1 - {\cos ^4}x = (1 - {\cos ^2}x)\,(1 + {\cos ^2}x)$

$ \Rightarrow $ ${\sin ^2}x[4{\sin ^2}x - 1 - (1 - {\sin ^2}x)] = 0$

$ \Rightarrow $${\sin ^2}x[5{\sin ^2}x - 2] = 0$

$ \Rightarrow $$\sin x = 0$ or $\sin x = \pm \sqrt {2/5} $.

Hence $x = n\pi $ is the required answer.

$ \Rightarrow $ $1 + \cos 3x + 1 + \sin \left( {2x - \frac{{7\pi }}{6}} \right) = 0$

$ \Rightarrow $ $(1 + \cos 3x) + 1 - \cos \left( {2x - \frac{{2\pi }}{3}} \right) = 0$

$ \Rightarrow $ $2{\cos ^2}\frac{{3x}}{2} + 2{\sin ^2}\left( {x - \frac{\pi }{3}} \right) = 0$

$ \Rightarrow $ $\cos \frac{{3x}}{2} = 0$ and $\sin \left( {x - \frac{\pi }{3}} \right) = 0$

$ \Rightarrow $$\frac{{3x}}{2} = \frac{\pi }{2},\,\frac{{3\pi }}{2},\,.....$

and $x - \frac{\pi }{3}$=0, $\pi ,2\pi ..... \Rightarrow x = \frac{\pi }{3}$

Therefore, the general solution of $\cos \frac{{3x}}{2} = 0$

and $\sin \left( {x - \frac{\pi }{3}} \right) = 0$ is $x = 2k\pi + \frac{\pi }{3} = \frac{\pi }{3}(6k + 1),$ where $k \in Z$.

==> $\theta = 2n\pi \pm \frac{\pi }{2}$.

$\tan \theta + \tan 2\theta + \sqrt 3 \tan \theta \tan 2\theta = \sqrt 3 $

==> $\,\tan \theta + \tan 2\theta = \sqrt 3 (1 - \tan \theta \tan 2\theta )$

==> $\frac{{\tan \theta + \tan 2\theta }}{{1 - \tan \theta \tan 2\theta }}$= $\sqrt 3 $

==> $\tan 3\theta = \tan (\pi /3)$

==> $3\theta = n\pi + \frac{\pi }{3}$

==> $\theta = (3n + 1)\frac{\pi }{9}$.

==> $2{\sin ^2}\frac{\theta }{2} = 2\sin \frac{\theta }{2}\,.\,\cos \frac{\theta }{2}\,.\,\sin \frac{\theta }{2}$

==> $2{\sin ^2}\frac{\theta }{2}\,\left[ {1 - \cos \frac{\theta }{2}} \right] = 0$

==> $\sin \frac{\theta }{2} = 0$ or $2{\sin ^2}\frac{\theta }{4} = 0$

==> $\sin \frac{\theta }{2} = 0$ or $\sin \frac{\theta }{4} = 0$

==> $\frac{\theta }{2} = k\pi $ or $\frac{\theta }{4} = k\pi $.

Hence, $\theta = 2k\pi $ or $\theta = 4k\pi $, $k \in I$.

$ \Rightarrow $ $ - \sin 3x = \sin 5x + \sin x = 2\sin 3x\cos 2x$

$ \Rightarrow $ $\sin 3x = 0$

$ \Rightarrow $ $x = 0$

or $\cos 2x = - \frac{1}{2} = - \cos \,\left( {\frac{\pi }{3}} \right) = \cos \,\left( {\pi - \frac{\pi }{3}} \right)$

$ \Rightarrow $ $2x = 2n\pi \pm \left( {\pi - \frac{\pi }{3}} \right)\, $

$\Rightarrow x = n\pi \pm \left( {\frac{\pi }{3}} \right)$

For $x$ lying between $0$ and $\frac{\pi }{2}$, we get $x = \frac{\pi }{3}$.

Trick : Check with options.

where $\cos \alpha = \frac{3}{5}$, $\sin \alpha = \frac{4}{5}$ 

Now $3\cos \theta + 4\sin \theta = k$  

$\therefore$ $5\cos (\theta - \alpha ) = k$

$\Rightarrow \cos (\theta - \alpha ) = \pm 1$

$ \Rightarrow $ $\theta - \alpha = {0^o},\,{180^o} $

$\Rightarrow \theta = \alpha ,\,{\rm{ }}{180^o} + \alpha $.

$ \Rightarrow $$2{\cos ^2}3\theta + 2\cos 3\theta \,.\,\cos \theta = 0$

$ \Rightarrow $ $4\cos 3\theta \cos 2\theta \cos \theta = 0$ 

$ \Rightarrow $ $3\theta = (2n + 1)\frac{\pi }{2};\,\,2\theta = (2n + 1)\frac{\pi }{2}{\rm{\, and\,\, }}\theta = (2n + 1)\frac{\pi }{2}$ 

$ \Rightarrow $ $\theta = {30^o},\,{90^o},\,{150^o},\,{45^o},\,{135^o}$.

or $\cos 2x(\cos 2x + 1) = 0$

$\therefore $ $\cos 2x = 0,\, - 1$, 

$\therefore $ $2x = \left( {n + \frac{1}{2}} \right){\rm{ }}\pi \,{\rm{\, or\,\,}}\,\,(2n + 1){\rm{ }}\pi $

$ \Rightarrow $ $x = (2n + 1)\frac{\pi }{4}{\rm{\,\, or\,\, }}(2n + 1)\frac{\pi }{2}$

Now put $n = - 2,\, - 1,\,0,\,1,\,2$

$\therefore $ $x = \frac{{ - 3\pi }}{4},\,\frac{{ - \pi }}{4},\,\frac{\pi }{4},\,\frac{{3\pi }}{4},\,\frac{{5\pi }}{4}$ 

and $\frac{{ - 3\pi }}{2},\,\frac{{ - \pi }}{2},\,\frac{\pi }{2},\,\frac{{3\pi }}{2},\,\frac{{5\pi }}{2}$

Since $ - \pi \le x \le \pi $, 

therefore $x \pm \frac{\pi }{4},\, \pm \frac{\pi }{2},\, \pm \frac{{3\pi }}{4}$ only.

$ \Rightarrow $ $2\sin 4\theta \cos 3\theta - \sin 4\theta = 0$

$ \Rightarrow $ $\sin 4\theta (2\cos 3\theta - 1) = 0 $

$\Rightarrow \sin 4\theta = 0,\,{\rm{ }}\cos 3\theta = \frac{1}{2}$

Now $\sin 4\theta = 0$ $ \Rightarrow $ $4\theta = \pi $ $ \Rightarrow $ $\theta = \frac{\pi }{4}$.

and $\cos 3\theta = \frac{1}{2}$ $ \Rightarrow $ $3\theta = \frac{\pi }{3}$ $ \Rightarrow $ $\theta = \frac{\pi }{9}$.

Then $\cos x = \frac{1}{2}{\rm{ as}}\,{\rm{ }}\cos x \ne \frac{{ - 3}}{2}$

$ \Rightarrow $ $x = 2n\pi \pm \frac{\pi }{3};\,\,\left\{ \begin{array}{l}{\rm{ \,for \,\,}}n = 0,\,\,x = \frac{\pi }{3},\,\frac{{5\pi }}{3}\\{\rm{ \,for\,\, }}n = 1,\,\,x = \frac{{5\pi }}{3}\end{array} \right\}$

and $\sin 2x + \cos 2x = \tan x$.....$(ii)$

Solving $(i)$, ${\sin ^2}2x = 2{\cos ^2}x$

==>$2{\cos ^2}x\cos 2x = 0$

==>$x = (2n + 1)\frac{\pi }{2}{\rm{ or }}x = (2n + 1)\frac{\pi }{4}$

$\therefore $ Common roots are $(2n \pm 1)\frac{\pi }{4}$

Solving $(ii)$, $\frac{{2\tan x + 1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}} = \tan x$

$ \Rightarrow $ ${\tan ^3}x + {\tan ^2}x - \tan x - 1 = 0$

$ \Rightarrow $ $({\tan ^2}x - 1)\,(\tan x + 1) = 0$

$ \Rightarrow $ $x = m\pi \pm \frac{\pi }{4}$

Trick : For $n = 0$, option $(a)$ gives $\theta = - \frac{\pi }{2}$ which satisfies the equation $(i)$ but does not satisfy the $(ii)$.

Now option $(b) $ gives $\theta = \frac{\pi }{4}$ which satisfies both the equations.

But $\sin \left( {2A - \frac{\pi }{6}} \right) - \frac{1}{2}$

attain maximum value at $2A - \frac{\pi }{6} = \frac{\pi }{2} \Rightarrow A = \frac{\pi }{3}$.

$\Rightarrow \frac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }} = 1$ 

$ \Rightarrow $ $\tan (\theta + \phi ) = 1$ 

$ \Rightarrow $ $\theta + \phi = \frac{\pi }{4} =  45^\circ$.

Dividing equation by $\sqrt 2 $,

$\frac{1}{{\sqrt 2 }}\cos x - \frac{1}{{\sqrt 2 }}\sin x = \frac{1}{2}$

$\cos \left( {\frac{\pi }{4} + x} \right) = \cos \frac{\pi }{3}$.

Hence, $\frac{\pi }{4} + x = 2n\pi \pm \frac{\pi }{3}$

$x = 2n\pi + \frac{\pi }{3} - \frac{\pi }{4} = 2n\pi + \frac{\pi }{{12}}$

or $x = 2n\pi - \frac{\pi }{3} - \frac{\pi }{4} = 2n\pi - \frac{{7\pi }}{{12}}$.

$12({\rm{cos}}{\rm{e}}{{\rm{c}}^2}\theta - 1) - 31{\rm{cos}}{\rm{ec }}\theta + {\rm{32}} = {\rm{0}}$

$12{\rm{cos}}{\rm{e}}{{\rm{c}}^2}\theta - 31\,{\rm{cos}}{\rm{ec }}\theta + {\rm{20}} = {\rm{0}}$

$12\,{\rm{cos}}{\rm{e}}{{\rm{c}}^2}\theta - 16\,\,{\rm{cos}}{\rm{ec }}\theta - 15{\rm{cos}}{\rm{ec}}\theta + 20 = {\rm{0}}$

$(4\cos {\rm{ec}}\theta - 5)(3\cos {\rm{ec}}\theta - 4) = 0$

${\rm{cos}}{\rm{ec}}\theta = \frac{5}{4},\frac{4}{3}$;

$\therefore$  $\sin \theta = \frac{4}{5},\frac{3}{4}$.

$2\cos 4\theta (\cos 3\theta + \cos \theta ) = 0$

==> $4\cos 4\theta \,\cos 2\theta \cos \theta = 0$

==> $4\frac{1}{{{2^3}\sin \theta }}$ $(\sin {2^3}\theta ) = 0$; $\sin 8\theta = 0$.

Hence $\theta = \frac{{n\pi }}{8}$.

$ \Rightarrow $ $\frac{1}{{1 - \sin x}} = 4 + 2\sqrt 3 $

$ \Rightarrow $ $\sin x = 1 - \frac{1}{{4 + 2\sqrt 3 }}$

$ \Rightarrow $ $\sin x = 1 - \frac{{(4 - 2\sqrt 3 )}}{4} = \frac{{2\sqrt 3 }}{4} = \frac{{\sqrt 3 }}{2}$

$ \Rightarrow $ $x = \frac{\pi }{3}$ or $\frac{{2\pi }}{3}$.

==> $p\theta = 2n\pi \pm (\pi + q\theta ),n \in I$

==> $\theta = \frac{{(2n + 1)\pi }}{{p - q}}$ or $\frac{{(2n - 1)\pi }}{{p + q}},\,\,n \in I$

Both the solutions form an $A.P.$ $\theta = \frac{{(2n + 1)\pi }}{{p - q}}$ gives us an $A.P.$ with common difference $\frac{{2\pi }}{{p - q}}$ and

$\theta = \frac{{(2n - 1)\pi }}{{p + q}}$ gives us an $A.P.$ with common difference $ = \frac{{2\pi }}{{p + q}}$. 

Certainly, $\frac{{2\pi }}{{p + q}} < \,\left| {\,\frac{{2\pi }}{{p - q}}\,} \right|$.

$ \Rightarrow $ $a\,\left( {\frac{{1 - {{\tan }^2}\,(x/2)}}{{1 + {{\tan }^2}(x/2)}}} \right) + \frac{{2b\tan (x/2)}}{{1 + {{\tan }^2}(x/2)}} = c$

$ \Rightarrow $ $(a + c){\tan ^2}\frac{x}{2} - 2b\tan \frac{x}{2} + (c - a) = 0$

This equation has roots $\tan \frac{\alpha }{2}$ and $\tan \frac{\beta }{2}$.

$\therefore $ $\tan \frac{\alpha }{2} + \tan \frac{\beta }{2} = \frac{{2b}}{{a + c}}$ and

$\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = \frac{{c - a}}{{a + c}}$

Now $\tan \left( {\frac{\alpha }{2} + \frac{\beta }{2}} \right) = \frac{{\tan \frac{\alpha }{2} + \tan \frac{\beta }{2}}}{{1 - \tan \frac{\alpha }{2}\tan \frac{\beta }{2}}}$

$= \frac{{\frac{{2b}}{{a + c}}}}{{1 - \frac{{c - a}}{{a + c}}}} = \frac{b}{a}$.

$\Rightarrow \quad x=\frac{2 n \pi}{3}, n \in I$

$\Rightarrow \quad x=\frac{2 \pi}{3}, \frac{4 \pi}{3} \in(0,2 \pi)$

Hence period of expression is $12\pi $ $(L.C.M.).$

$ \Rightarrow $ $L.C.M.$ is $\pi $.