Exercise (11.2) No.1 Solution


(a) $ \displaystyle \ \ \ \frac{{\tan \theta -1+\sec \theta }}{{\tan \theta +1-\sec \theta }}$

$ \displaystyle \text{=}\frac{{\tan \theta +\sec \theta -1}}{{\tan \theta -\sec \theta +1}}$

$ \displaystyle \text{=}\frac{{\left( {\tan \theta +\sec \theta } \right)-\left( {{{{\sec }}^{2}}\theta -{{{\tan }}^{2}}\theta } \right)}}{{\tan \theta -\sec \theta +1}}\ $
$ \displaystyle \ \ \ \left[ {\because {{{\tan }}^{2}}\theta +1={{{\sec }}^{2}}\theta } \right]$

$ \displaystyle \text{=}\frac{{\left( {\tan \theta +\sec \theta } \right)-\left( {\sec \theta +\tan \theta } \right)\left( {\sec \theta -\tan \theta } \right)}}{{\tan \theta +1-\sec \theta }}$

$ \displaystyle \text{=}\frac{{\left( {\tan \theta +\sec \theta } \right)\left( {1-\sec \theta +\tan \theta } \right)}}{{1-\sec \theta +\tan \theta }}$

$ \displaystyle =\tan \theta +\sec \theta $

$ \displaystyle =\frac{{\sin \theta }}{{\cos \theta }}+\frac{1}{{\cos \theta }}$

$ \displaystyle =\frac{{1+\sin \theta }}{{\cos \theta }}$


(b) $ \displaystyle \ \ \ \frac{{1-\cos x}}{{1+\cos x}}$

$ \displaystyle =\frac{{1-\cos x}}{{1+\cos x}}\times \frac{{1-\cos x}}{{1-\cos x}}$

$ \displaystyle =\frac{{{{{\left( {1-\cos x} \right)}}^{2}}}}{{1-{{{\cos }}^{2}}x}}$

$ \displaystyle =\frac{{1-2\cos x+{{{\cos }}^{2}}x}}{{{{{\sin }}^{2}}x}}$

$ \displaystyle =\frac{1}{{{{{\sin }}^{2}}x}}-\frac{2}{{\sin x}}\cdot \frac{{\cos x}}{{\sin x}}+\frac{{{{{\cos }}^{2}}x}}{{{{{\sin }}^{2}}x}}$

$ \displaystyle ={{\operatorname{cosec}}^{2}}x-2\operatorname{cosec}x\cot x+{{\cot }^{2}}x$

$ \displaystyle ={{\operatorname{cosec}}^{2}}x-2\operatorname{cosec}x\cot x+{{\operatorname{cosec}}^{2}}x-1\ $
$ \displaystyle \ \ \ \left[ {\because 1+{{{\cot }}^{2}}x={{{\operatorname{cosec}}}^{2}}x} \right]$

$ \displaystyle =2{{\operatorname{cosec}}^{2}}x-2\operatorname{cosec}x\cot x-1$


(c)$ \displaystyle \ \ \ (1+\tan x-\sec x)(1+\cot x+\operatorname{cosec}x)$

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

$ \displaystyle =\frac{{\cos x+\sin x-1}}{{\cos x}}\ \times \frac{{\sin x+\cos x+1}}{{\sin x}}$

$ \displaystyle =\frac{{{{{(\cos x+\sin x)}}^{2}}-1}}{{\sin x\cos x}}$

$ \displaystyle =\frac{{{{{\cos }}^{2}}x+2\sin x\cos x+{{{\sin }}^{2}}x-1}}{{\sin x\cos x}}$

$ \displaystyle =\frac{{{{{\cos }}^{2}}x+{{{\sin }}^{2}}x+2\sin x\cos x-1}}{{\sin x\cos x}}$

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

$ \displaystyle =2$