$B=\dfrac{sinx + tanx}{tanx} - sinx.cosx$ 

$= \dfrac{sinx}{tanx} +1-sinx.cosx$ 

$= cosx +1-sinx.cosx$ 

$C=tanx +\dfrac{cosx}{1+sinx}$

$= \dfrac{sinx}{cosx} + \dfrac{cosx}{1+sinx}$ 

$= \dfrac{sinx. (1+sinx)+cos^{2}x}{cosx.(1+sinx)}$ 

$= \dfrac{sinx + sin^{2}x +cos^{2}x}{cosx l. (1+sinx)}$ 

$= \dfrac{sinx +1}{cosx. (1+sinx)}$ 

$= \dfrac{1}{cosx}$ 

$D= \dfrac{cosx.tanx}{sin^{2}x} - cosx.cotx$ 

$= \dfrac{sinx}{sin^{2}x} -  cosx. cotx$

$=\dfrac{1}{sinx} -  \dfrac{cos^{2}x}{sinx}$ 

$= \dfrac{1-cos^{2}x}{sinx} = sinx$ 

$E= (1+sinx).tan^{2}x.(1-sinx)$ 

$= (1+sinx). (1-sinx).tan^{2}x$

$= (1-sin^{2}x).tan^{2}x$ 

$= cos^{2}x. \dfrac{sin^{2}x}{cos^{2}x}$

$= sin^{2}x$ 

$F= 1- \dfrac{sin^{2}x}{1+cotx} - \dfrac{cos^{2}x}{1+tanx}$ 

$=1-\dfrac{sin^{2}x}{1+\dfrac{cosx}{sinx}} - \dfrac{cos^{2}x}{1+\dfrac{sinx}{cosx}}$ 

$=1-\dfrac{sin^{3}x}{cosx +sinx} - \dfrac{cos^{3}x}{sinx + cosx}$ 

$= 1-\dfrac{sin^{3}x +cos^{3}x}{sinx +cosx}$ 

$=1-(sin^{2}x- sinx.cosx +cos^{2}x)$

$= sinx.cosx$ 

$\begin{array}{l}+) \quad B = \dfrac{\sin x + \tan x}{\tan x} - \sin x.\cos x \cr \to B = \dfrac{\sin x}{\tan x} + 1 - \sin x.\cos x\cr \to B = \dfrac{\sin x}{\dfrac{\sin x}{\cos x}}+ 1 - \sin x.\cos x\cr \to B = \cos x + 1 - \sin x.\cos x\\ +) \quad C = \tan x + \dfrac{\cos x}{1+\sin x}\\ \to C = \dfrac{\sin x}{\cos x}+ \dfrac{\cos x}{1+\sin x}\\ \to C = \dfrac{\sin x(1 + \sin x) + \cos^2x}{\cos x(1 + \sin x)}\\ \to C = \dfrac{\sin x + \sin^2x + \cos^2x}{\cos x(1 + \sin x)}\\ \to C = \dfrac{\sin x + 1}{\cos x(1 + \sin x)}\\ \to C = \dfrac{1}{\cos x}\\ +) \quad D = \dfrac{\cos x.\tan x}{\sin^2x} -\cos x.\cot x\\ \to D = \dfrac{\cos x\cdot \dfrac{\sin x}{\cos x}}{\sin^2x} - \cos x\cdot\dfrac{\cos x}{\sin x} \\ \to D = \dfrac{1}{\sin x} - \dfrac{\cos^2x}{\sin x} \\ \to D = \dfrac{1 - \cos^2x}{\sin x}\\ \to D = \dfrac{\sin^2x}{\sin x}\\ \to D = \sin x\\ +) \quad E = (1+\sin x).\tan^2x.(1 - \sin x)\\ \to E = (1 - \sin^2x).\tan^2x\\ \to E = \cos^2x\cdot\dfrac{\sin^2x}{\cos^2x}\\ \to E = \sin^2x\\ +) \quad F = 1 - \dfrac{\sin^2x}{1 + \cot x} - \dfrac{\cos^2x}{1 + \tan x}\\ \to F = \left(\sin^2x -\dfrac{\sin^2x}{1 + \cot x}\right) + \left(\cos^2x -- \dfrac{\cos^2x}{1 + \tan x}\right)\\ \to F = \sin^2x\cdot\left( 1 - \dfrac{1}{1 + \cot x}\right) + \cos^2x\cdot\left( 1 - \dfrac{1}{1 + \tan x}\right)\\ \to F = \sin^2x\cdot\dfrac{1 + \cot x - 1}{1 + \cot x} + \cos^2x\cdot\dfrac{1 + \tan x - 1}{1 + \tan x}\\ \to F = \dfrac{\sin^2x.\cot x}{1 + \cot x} + \dfrac{\cos^2x.\tan x}{1 + \tan x}\\ \to F = \dfrac{\sin^2x.\dfrac{\cos x}{\sin x}}{1 + \dfrac{\cos x}{\sin x}} + \dfrac{\cos^2x.\dfrac{\sin x}{\cos x}}{1 + \dfrac{\sin x}{\cos x}}\\ \to F = \dfrac{\sin^2x.\cos x}{\sin x + \cos x} + \dfrac{\cos^2x.\sin x}{\sin x + \cos x}\\ \to F = \dfrac{\sin^2x.\cos x + \cos^2x.\sin x}{\sin x + \cos x}\\ \to F = \dfrac{\sin x.\cos x(\sin x + \cos x)}{\sin x + \cos x}\\ \to F = \sin x.\cos x\end{array}$