Ans. $$(a)$$ 
Since $$0<\alpha <\pi$$ we have $$\sin { \alpha  }$$ =$$\dfrac{12}{13}$$
$$\Rightarrow \cos { \alpha  } =\surd \left[ 1-{ \left( 12/13 \right)  }^{ 2 } \right] $$=$$\dfrac{5}{13}$$
And since $$\pi <\beta <3\pi /2$$, we have $$\cos { \beta  } =-3/5\Rightarrow \sin { \beta  } =-4/5$$
Hence $$\sin { \left( \alpha +\beta  \right)  } =\sin { \alpha  } \cos { \beta  } +\cos { \alpha  } \sin { \beta  }$$ 
$$=\dfrac { 12 }{ 13 } \left( -\dfrac { 3 }{ 5 }  \right) +\left( \dfrac { 5 }{ 13 }  \right) \left( -\dfrac { 4 }{ 5 }  \right) =-\dfrac { 56 }{ 65 } $$

Ans. Ans. $$(a)$$.
$$f\left( \theta  \right) =\dfrac { 1 }{ \sin ^{ 2 }{ \theta  } +3\sin \theta \cos \theta +5\cos ^{ 2 }{ \theta  }  }$$
$$=\dfrac { 1 }{ \dfrac { 1-\cos { 2\theta  }  }{ 2 } +\dfrac { 3 }{ 2 } \sin { 2\theta  } +\dfrac { 5\left( 1+\cos { 2\theta  }  \right)  }{ 2 }  } $$
$$=\dfrac{ 2 }{ 6+3\sin 2\theta+4 \cos 2\theta }$$
Hence, $$\left[ f\left( \theta  \right)  \right] _{ maximum }=\dfrac { 2 }{ 6-5 } =2$$

Ans. $$(a)$$ and $$(c)$$
$$\cos { \left( A-B \right) = } 3/5$$
$$\therefore\quad 5\cos { A } \cos { B } +5\sin { A } \sin { B } =3$$
But from $$2$$nd relation 
$$\sin { A } \sin { B } =2\cos { A } \cos { B }$$
$$\therefore\quad \left( 5+10 \right) \cos { A } \cos { B } =3$$
$$\therefore\quad \cos { A } \cos { B } =1/5$$ i.e. $$(a)$$ 
Or $$5\left( \dfrac { 1 }{ 2 } +1 \right) \sin { A } \sin { B } =3$$
$$\therefore\quad \sin { A } \sin { B } =2/5$$ $$\quad\therefore\quad $$ $$(b)$$ is not correct.
$$\cos { \left( A+B \right)  } =\cos { A } \cos { B } -\sin { A } \sin { B } =\cfrac { 1 }{ 5 } -\cfrac { 2 }{ 5 } =-\cfrac { 1 }{ 5 }$$ i.e. $$(c)$$

From $$1$$st, $$2\sin^2\theta -(1-2\sin^2\theta)=0$$
$$\therefore 4\sin^2\theta =1$$ or $$\sin\theta =\pm 1/2$$
$$\therefore \theta =\pi/6, \pi +\pi/6$$ in $$[0, 2\pi]$$
Also from $$2$$nd equation, we have
$$2(1-\sin^2\theta)-3\sin\theta =0$$
or $$2\sin^2\theta +3\sin\theta -2=0$$
or $$(\sin \theta +2)(2\sin \theta -1)=0$$
$$\therefore \sin\theta =1/2$$ as $$-2$$ is rejected
$$\therefore \sin\theta =1/2$$ which in already included in $$1$$st. Hence there are only two solutions $$\Rightarrow$$ (c).

We have $$\sin x - \sin\lambda$$
$$\Rightarrow    x = n\pi + (-1)^{n} \lambda,n\epsilon l$$
$$\therefore$$ $$\sin (x/3) = \sin [n\pi/3 + (- 1)^{2} (X/3)]$$
Hence for n = 0,3, we have
$$\sin(x/3) = \sin(\lambda/3)$$
and for n = 1,2, we have
$$\sin (x /3)$$ = $$\sin{\pi/3-\lambda/3}$$
[Note that for n = 2, sin(x /3)
=$$ \sin (2\pi /3 + \lambda/3)$$ = $$\sin [\pi - [\pi /3 - \lambda/3]]$$
= $$\sin(\pi/3-\lambda/3)$$
and for n = 4,5, we get
sin (x/3) = -$$\sin(\pi/3 + \lambda/3)$$.
It is easy to see that all other values of n repeat only these values of $$\sin (x/3)$$.

Ans. (b)
$$\sin\alpha =\lambda$$
$$\theta =n\pi +(-1)^n\alpha$$
For $$+$$ive values, $$n=0, 1, 2, 3$$
$$\therefore \alpha =\alpha, \beta =\pi -\alpha, \gamma =2\pi +\alpha , \delta =3\pi -\alpha$$
$$\therefore E=4\sin\dfrac{\alpha}{2}+3\sin\left(\dfrac{\pi}{2}-\dfrac{\alpha}{2}\right)+2\sin\left(\pi +\dfrac{\alpha}{2}\right)+\sin\left(\dfrac{3\pi}{2}-\dfrac{\alpha}{2}\right)$$
$$=4\sin\dfrac{\alpha}{2}+3\cos\dfrac{\alpha}{2}-2\sin\dfrac{\alpha}{2}-\cos\dfrac{\alpha}{2}$$
$$=2\left(\cos\dfrac{\alpha}{2}+\sin\dfrac{\alpha}{2}\right)=2\sqrt{\left(\cos \dfrac{\alpha}{2}+\sin\dfrac{\alpha}{2}\right)^2}$$
$$=2\sqrt{1+\sin\alpha}=2\sqrt{1+\lambda}$$.

Put $$\cos^2x=1-\sin^2x$$
If $$y=81^{\sin^2x}$$ then $$y+\dfrac{81}{y}=30$$
$$y^2-30y+81=0$$, $$y=27, 3$$
$$3^{4\sin^2x}=3^3, 3^1$$
$$\therefore \sin^2x=\dfrac{3}{4}, \dfrac{1}{4}$$
$$\therefore \sin x=\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}$$
as $$\sin x$$ is $$+$$ive in the given interval $$0 < x < \pi$$
$$\therefore x=\dfrac{\pi}{3}, \pi -\dfrac{\pi}{3}=\dfrac{2\pi}{3}, \dfrac{\pi}{6}, \pi -\dfrac{\pi}{6}=\dfrac{5\pi}{6}$$.

$$2(\cos x+2\cos^2 x-1)+2\sin x\cos x. (1+2\cos x)-2\sin x=0$$
or $$2(2\cos^2x+\cos x-1)+2\sin x(2\cos^2x+\cos x-1)=0$$
$$2(1+\sin x)(\cos x+1)(2\cos x-1)=0$$
We have to determine values of x s.t. $$-\pi \leq x\leq \pi$$
$$1+\sin x=0$$ $$\therefore \sin x=-1$$
$$\therefore x=2n\pi +\dfrac{3\pi}{2}$$  $$\therefore x=-\dfrac{\pi}{2}$$,
for $$n=-1$$ $$\because -\pi \leq x < \pi$$
$$\cos x=-1=\cos \pi$$ $$\therefore \cos x=1/2=\cos (\pi/3)$$
$$\therefore x=2n\pi \pm \pi/3$$
$$\therefore x=\pi /3$$, $$-\pi /3$$
Hence the values of x s.t. $$-\pi \leq x \leq \pi$$ are
$$-\pi, -\pi /2, -\pi /3, \pi/3, \pi$$.

$$(2+\sqrt{3})\cos\theta =1-\sin\theta$$
$$\dfrac{1-\sin\theta}{\cos\theta}=2+\sqrt{3}$$
$$\dfrac{\left(\cos\dfrac{\theta}{2}-\sin\dfrac{\theta}{2}\right)^2}{\cos^2\dfrac{\theta}{2}-\sin^2\dfrac{\theta}{2}}=2+\sqrt{3}$$
or $$\dfrac{1-t}{1+t}=\tan 75^o, t=\tan \dfrac{\theta}{2}$$
or $$\tan\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)=\tan\dfrac{5\pi}{12}$$
$$\therefore \dfrac{\pi}{4}-\dfrac{\theta}{2}=n\pi +\dfrac{5\pi}{12}$$
or $$\theta =-2n\pi -\dfrac{2\pi}{3}$$
or $$\theta =2r\pi -\dfrac{2\pi}{3}, r\in I$$.

$$x = h \cot 3 \alpha $$                          ...............(i)

$$\tan\theta +sec\theta =\sqrt{3}$$
$$\dfrac{1+\sin\theta}{\cos\theta}=\sqrt{3}$$
$$\dfrac{\left(\cos\dfrac{\theta}{2}+\sin\dfrac{\theta}{2}\right)^2}{\cos^2\dfrac{\theta}{2}-\sin^2\dfrac{\theta}{2}}=\sqrt{3}$$
or $$\dfrac{\cos\dfrac{\theta}{2}+\sin\dfrac{\theta}{2}}{\cos\dfrac{\theta}{2}-\sin\dfrac{\theta}{2}}=\sqrt{3}$$ or $$\dfrac{1+t}{1-t}=\sqrt{3}$$
or $$\tan\left(\dfrac{\pi}{4} +\dfrac{\theta}{2}\right)=\tan\dfrac{\pi}{3}$$
$$\therefore \dfrac{\theta}{2}+\dfrac{\pi}{4}=n\pi +\dfrac{\pi}{3}$$
$$\therefore \theta =2n\pi +\dfrac{\pi}{6}$$.

$$secA+tanA=m$$ ......(1)

$$\sec^2\theta=1+\tan^2\theta$$

$$\begin{array}{l} f\left( x \right) =a\sin  x+\frac { 1 }{ 3 } \sin  3x \\ { f^{ ' } }\left( x \right) =a\cos  x+\dfrac { 1 }{ 3 } \cos  3x\times 3 \\ =a\cos  x+\cos  3x \\ Since\, f\left( x \right) has\, \max  imum\, at\, x=\dfrac { \pi  }{ 3 }  \\ so,\, f'\left( { \dfrac { \pi  }{ 3 }  } \right) =0 \\ \Rightarrow a\cos  \dfrac { \pi  }{ 3 } +\cos  3\times \dfrac { \pi  }{ 3 } =0 \\ \Rightarrow a\times \dfrac { 1 }{ 2 } +\left( { -1 } \right) =0 \\ \Rightarrow \dfrac { a }{ 2 } =1 \\ a=2 \end{array}$$

$$\alpha\cos^2 3\theta+\beta\cos^4\theta=16\cos^6\theta+9\cos^2\theta$$

We have,

$$\implies\cfrac { 1-\cos A }{ 2 } =x$$

$$\sin\theta =n\sin(\theta +2\alpha )$$

Given equation $$3x+2\tan x=\dfrac{5\pi}{2}$$