## Monday, January 21, 2019

### Sines Law , Componendo and Dividendo

ABC is a triangle having  BC = 2 AB . Bisect  BC  in  D  and  BD  in  E .
Prove that  AD  bisects  $\displaystyle \angle \ CAE.$

Solution

$\displaystyle \text{Let}\ AB=2x$

$\displaystyle \ \ \ \therefore \ BC=4x\ \ \ \left( {\because \ BC=2AB} \right)$

$\displaystyle \text{Bisect}\ BC\ \ \text{in}\ \ D\ \ \text{and}\ \ BD\ \ \text{in}\ \ E\ .\ \ \left( {\text{Given}} \right)$

$\displaystyle BD=DC=2x\ and\ \ BE=ED=x$

By the law of sines ,

$\displaystyle \frac{{AB}}{{\sin \gamma }}=\frac{{BC}}{{\sin \left( {\phi +\beta +\alpha } \right)}}$

$\displaystyle \frac{{2x}}{{\sin \left( {\theta -\alpha } \right)}}=\frac{{4x}}{{\sin \left( {\theta +\alpha } \right)}}$

$\displaystyle \frac{{\sin \left( {\theta +\alpha } \right)}}{{\sin \left( {\theta -\alpha } \right)}}=2\ \ \ \ ...........(1)$

$\displaystyle \frac{{BE}}{{\sin \phi }}=\frac{{AB}}{{\sin \lambda }}$

$\displaystyle \frac{x}{{\sin \left( {\theta -\beta } \right)}}=\frac{{2x}}{{\sin \left( {\theta +\beta } \right)}}$

$\displaystyle \frac{{\sin \left( {\theta +\beta } \right)}}{{\sin \left( {\theta -\beta } \right)}}=2\ \ \ \ ...........(2)$

$\displaystyle \text{From}\ (1)\ \text{and}\ (2)\ ,$

$\displaystyle \frac{{\sin \left( {\theta +\alpha } \right)}}{{\sin \left( {\theta -\alpha } \right)}}=\frac{{\sin \left( {\theta +\beta } \right)}}{{\sin \left( {\theta -\beta } \right)}}$

$\displaystyle \text{By}\ \text{Componendo}\ \And \ \text{Dividendo}\ ,$

$\displaystyle \frac{{\sin \left( {\theta +\alpha } \right)+\sin \left( {\theta -\alpha } \right)}}{{\sin \left( {\theta +\alpha } \right)-\sin \left( {\theta -\alpha } \right)}}=\frac{{\sin \left( {\theta +\beta } \right)+\sin \left( {\theta -\beta } \right)}}{{\sin \left( {\theta +\beta } \right)-\sin \left( {\theta -\beta } \right)}}$

$\displaystyle \frac{{2\sin \theta \cos \alpha }}{{2\cos \theta \sin \alpha }}=\frac{{2\sin \theta \cos \beta }}{{2\cos \theta \sin \beta }}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \cot \alpha =\cot \beta$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \alpha =\beta$

$\displaystyle \therefore \ AD\ \text{bisects}\ \angle \ CAE.$

### Solving problem with compound angle formula

$\displaystyle A+B+C={{180}^{\circ }}$

$\displaystyle 3\sin A+4\cos B=6$

$\displaystyle 3\cos A+4\sin B=\sqrt{{13}}$

$\displaystyle \sin x=?$

$\displaystyle \text{Solution}$

$\displaystyle A+B+C={{180}^{\circ }}$

$\displaystyle 3\sin A+4\cos B=6$

$\displaystyle {{\left( {3\sin A+4\cos B} \right)}^{2}}={{6}^{2}}$

$\displaystyle 9{{\sin }^{2}}A+24\sin A\cos B+16{{\cos }^{2}}B=36\ ..........(1)$

$\displaystyle 3\cos A+4\sin B=\sqrt{{13}}$

$\displaystyle {{\left( {3\cos A+4\sin B} \right)}^{2}}={{\left( {\sqrt{{13}}} \right)}^{2}}$

$\displaystyle 9{{\cos }^{2}}A+24\cos A\sin B+16{{\sin }^{2}}B=13\ ............(2)$

$\displaystyle (1)+(2)\Rightarrow \$

$\displaystyle 9\left( {{{{\sin }}^{2}}A+{{{\cos }}^{2}}A} \right)+24\left( {\sin A\cos B+\cos A\sin B} \right)+16\left( {{{{\sin }}^{2}}B+{{{\cos }}^{2}}B} \right)=49$

$\displaystyle \ 9+24\sin \left( {A+B} \right)+16=49$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ 24\sin \left( {A+B} \right)=24$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sin \left( {A+B} \right)=1$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \sin \left( {{{{180}}^{\circ }}-C} \right)=1$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \ \ \sin C=1$

### Solving problem with Factor and sum formulae

$\displaystyle \text{Prove}\ \text{that}\ \ \ \frac{{\cos 8x-\cos 7x}}{{1+2\cos 5x}}=\cos 3x-\cos 2x\ .$

Solution

$\displaystyle \ \ \ \frac{{\cos 8x-\cos 7x}}{{1+2\cos 5x}}=\frac{{-2\sin \frac{{15x}}{2}\sin \frac{x}{2}}}{{1+2\left( {2{{{\cos }}^{2}}\frac{{5x}}{2}-1} \right)}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{-2\sin \frac{{15x}}{2}\sin \frac{x}{2}}}{{4{{{\cos }}^{2}}\frac{{5x}}{2}-1}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{-2\sin \frac{{15x}}{2}\sin \frac{x}{2}}}{{4{{{\cos }}^{2}}\frac{{5x}}{2}-1}}\ \ \times \ \ \frac{{-2{{{\sin }}^{2}}\frac{{5x}}{2}}}{{-2{{{\sin }}^{2}}\frac{{5x}}{2}}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\left( {-2\sin \frac{{5x}}{2}\sin \frac{x}{2}} \right)\left( {-2\sin \frac{{15x}}{2}\sin \frac{{5x}}{2}} \right)}}{{2{{{\sin }}^{2}}\frac{{5x}}{2}-8{{{\sin }}^{2}}\frac{{5x}}{2}{{{\cos }}^{2}}\frac{{5x}}{2}}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\left( {\cos 3x-\cos 2x} \right)\left( {\cos 10x-\cos 5x} \right)}}{{2{{{\sin }}^{2}}\frac{{5x}}{2}-2{{{\left( {2\sin \frac{{5x}}{2}\cos \frac{{5x}}{2}} \right)}}^{2}}}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\left( {\cos 3x-\cos 2x} \right)\left( {\cos 10x-\cos 5x} \right)}}{{2{{{\sin }}^{2}}\frac{{5x}}{2}-2{{{\sin }}^{2}}5x}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\left( {\cos 3x-\cos 2x} \right)\left( {\cos 10x-\cos 5x} \right)}}{{\left( {1-\cos 5x} \right)-\left( {1-\cos 10x} \right)}}\ \ \ \ \ \ \left[ {\because \cos 2\theta =1-2{{{\sin }}^{2}}\theta } \right]$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\left( {\cos 3x-\cos 2x} \right)\left( {\cos 10x-\cos 5x} \right)}}{{\cos 10x-\cos 5x}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\cos 3x-\cos 2x$

### Solving problem with Sines law and Cosines law

In the diagram below , the lengths of the three sides of the triangle are $\displaystyle a\ cm\ ,\ b\ cm\ \text{and}\ c\ cm$ . It is given that $\displaystyle \frac{{{{a}^{2}}+{{b}^{2}}}}{{{{c}^{2}}}}=2011$ . Find the value of $\displaystyle \frac{{\cot C}}{{\cot A\ +\ \cot B}}$ .

Solution

$\displaystyle \frac{{{{a}^{2}}+{{b}^{2}}}}{{{{c}^{2}}}}=2011$

$\displaystyle {{a}^{2}}+{{b}^{2}}=2011\ {{c}^{2}}$

$\displaystyle {{a}^{2}}+{{b}^{2}}-{{c}^{2}}=2010\ {{c}^{2}}$

$\displaystyle \frac{{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}}{{2ab}}=\frac{{2010\ {{c}^{2}}}}{{2ab}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \cos C=\frac{{2010\ {{c}^{2}}}}{{2ab}}\ \ \ \ \ \ \ \ \ \ \ \ \ \left[ {\because \cos C=\frac{{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}}{{2ab}}} \right]$

$\displaystyle \ \ \ \ \ \ \ \ \ \cos C=1005\left( {\frac{c}{a}} \right)\left( {\frac{c}{b}} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \cos C=1005\left( {\frac{{\sin C}}{{\sin A}}} \right)\left( {\frac{{\sin C}}{{\sin B}}} \right)\ \ \ \left[ {\because \ \frac{a}{{\sin A}}=\frac{b}{{\sin B}}=\frac{c}{{\sin C}}} \right]$

$\displaystyle \ \ \ \ \ \ \ \ \ \frac{{\cos C}}{{\sin C}}=\frac{{1005\sin C}}{{\sin A\ \sin B}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \cot C=\frac{{1005\sin \left( {{{{180}}^{\circ }}-\left( {A+B} \right)} \right)}}{{\sin A\ \sin B}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \cot C=\frac{{1005\sin \left( {A+B} \right)}}{{\sin A\ \sin B}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \cot C=1005\left( {\frac{{\sin A\ \cos B+\cos A\ \sin B}}{{\sin A\ \sin B}}} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \cot C=1005\left( {\frac{{\sin A\ \cos B}}{{\sin A\ \sin B}}+\frac{{\cos A\ \sin B}}{{\sin A\ \sin B}}} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \cot C=1005\left( {\cot B+\cot A} \right)$

$\displaystyle \therefore \ \frac{{\cot C}}{{\cot A\ +\ \cot B}}=1005$

### Solving problem with double angle formula and factor formula

Find the value of
$\displaystyle (\ \text{i}\ )\ \ \ {{\sin }^{4}}\frac{\pi }{8}+{{\cos }^{4}}\frac{\pi }{8}+{{\sin }^{4}}\frac{{7\pi }}{8}+{{\cos }^{4}}\frac{{7\pi }}{8}$

$\displaystyle (\ ii\ )\ \frac{3}{{{{{\sin }}^{2}}{{{20}}^{\circ }}}}-\frac{1}{{{{{\cos }}^{2}}{{{20}}^{\circ }}}}+64{{\sin }^{2}}{{20}^{\circ }}$

Solution

$\displaystyle (\ \text{i}\ )\ \ \ {{\sin }^{4}}\frac{\pi }{8}+{{\cos }^{4}}\frac{\pi }{8}+{{\sin }^{4}}\frac{{7\pi }}{8}+{{\cos }^{4}}\frac{{7\pi }}{8}$

$\displaystyle \ \ \ \ \ ={{\left( {{{{\sin }}^{2}}\frac{\pi }{8}+{{{\cos }}^{2}}\frac{\pi }{8}} \right)}^{2}}-2{{\sin }^{2}}\frac{\pi }{8}{{\cos }^{2}}\frac{\pi }{8}+{{\left( {{{{\sin }}^{2}}\frac{{7\pi }}{8}+{{{\cos }}^{2}}\frac{{7\pi }}{8}} \right)}^{2}}-2{{\sin }^{2}}\frac{{7\pi }}{8}{{\cos }^{2}}\frac{{7\pi }}{8}$

$\displaystyle \ \ \ \ \ =1-\frac{1}{2}{{\left( {2\sin \frac{\pi }{8}\cos \frac{\pi }{8}} \right)}^{2}}+1-\frac{1}{2}{{\left( {2\sin \frac{{7\pi }}{8}\cos \frac{{7\pi }}{8}} \right)}^{2}}$

$\displaystyle \ \ \ \ \ =2-\frac{1}{2}{{\sin }^{2}}\frac{\pi }{4}-\frac{1}{2}{{\sin }^{2}}\frac{{7\pi }}{4}$

$\displaystyle \ \ \ \ \ =2-\frac{1}{2}\left( {{{{\sin }}^{2}}\frac{\pi }{4}+{{{\sin }}^{2}}\frac{{7\pi }}{4}} \right)$

$\displaystyle \ \ \ \ \ =2-\frac{1}{2}\left( {{{{\sin }}^{2}}\frac{\pi }{4}+{{{\sin }}^{2}}\left( {2\pi -\frac{\pi }{4}} \right)} \right)$

$\displaystyle \ \ \ \ \ =2-\frac{1}{2}\left( {{{{\sin }}^{2}}\frac{\pi }{4}+{{{\sin }}^{2}}\frac{\pi }{4}} \right)\ \ \ \ \left[ {\because \sin \left( {2\pi -\theta } \right)=-\sin \theta \ ,\ {{{\sin }}^{2}}\left( {2\pi -\theta } \right)={{{\sin }}^{2}}\theta } \right]$

$\displaystyle \ \ \ \ \ =2-{{\sin }^{2}}\frac{\pi }{4}$

$\displaystyle \ \ \ \ \ =2-\frac{1}{2}=\frac{3}{2}$

$\displaystyle (\ ii\ )\ \frac{3}{{{{{\sin }}^{2}}{{{20}}^{\circ }}}}-\frac{1}{{{{{\cos }}^{2}}{{{20}}^{\circ }}}}+64{{\sin }^{2}}{{20}^{\circ }}$

$\displaystyle \ \ \ \ =\frac{{3{{{\cos }}^{2}}{{{20}}^{\circ }}-{{{\sin }}^{2}}{{{20}}^{\circ }}+64{{{\sin }}^{4}}{{{20}}^{\circ }}{{{\cos }}^{2}}{{{20}}^{\circ }}}}{{{{{\sin }}^{2}}{{{20}}^{\circ }}{{{\cos }}^{2}}{{{20}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{3\left( {1-{{{\sin }}^{2}}{{{20}}^{\circ }}} \right)-{{{\sin }}^{2}}{{{20}}^{\circ }}+16{{{\sin }}^{2}}{{{20}}^{\circ }}\left( {4{{{\sin }}^{2}}{{{20}}^{\circ }}{{{\cos }}^{2}}{{{20}}^{\circ }}} \right)}}{{\frac{1}{4}\left( {4{{{\sin }}^{2}}{{{20}}^{\circ }}{{{\cos }}^{2}}{{{20}}^{\circ }}} \right)}}$

$\displaystyle \ \ \ \ =\frac{{3-3{{{\sin }}^{2}}{{{20}}^{\circ }}-{{{\sin }}^{2}}{{{20}}^{\circ }}+16{{{\sin }}^{2}}{{{20}}^{\circ }}{{{\left( {2\sin {{{20}}^{\circ }}\cos {{{20}}^{\circ }}} \right)}}^{2}}}}{{\frac{1}{4}{{{\left( {2\sin {{{20}}^{\circ }}\cos {{{20}}^{\circ }}} \right)}}^{2}}}}$

$\displaystyle \ \ \ \ =\frac{{3-4{{{\sin }}^{2}}{{{20}}^{\circ }}+16{{{\sin }}^{2}}{{{20}}^{\circ }}{{{\sin }}^{2}}{{{40}}^{\circ }}}}{{\frac{1}{4}{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{3-4{{{\sin }}^{2}}{{{20}}^{\circ }}+{{{\left( {4\sin {{{40}}^{\circ }}\sin {{{20}}^{\circ }}} \right)}}^{2}}}}{{\frac{1}{4}{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{3-4{{{\sin }}^{2}}{{{20}}^{\circ }}+{{{\left( {2\left( {\cos {{{20}}^{\circ }}-\cos {{{60}}^{\circ }}} \right)} \right)}}^{2}}}}{{\frac{1}{4}{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{3-4{{{\sin }}^{2}}{{{20}}^{\circ }}+4{{{\left( {\cos {{{20}}^{\circ }}-\cos {{{60}}^{\circ }}} \right)}}^{2}}}}{{\frac{1}{4}{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{12-16{{{\sin }}^{2}}{{{20}}^{\circ }}+16{{{\left( {\cos {{{20}}^{\circ }}-\frac{1}{2}} \right)}}^{2}}}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{12-16{{{\sin }}^{2}}{{{20}}^{\circ }}+16\left( {{{{\cos }}^{2}}{{{20}}^{\circ }}-\cos {{{20}}^{\circ }}+\frac{1}{4}} \right)}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{12-16{{{\sin }}^{2}}{{{20}}^{\circ }}+16{{{\cos }}^{2}}{{{20}}^{\circ }}-16\cos {{{20}}^{\circ }}+4}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{16\left( {1+{{{\cos }}^{2}}{{{20}}^{\circ }}-{{{\sin }}^{2}}{{{20}}^{\circ }}-\cos {{{20}}^{\circ }}} \right)}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{16\left( {1+\cos {{{40}}^{\circ }}-\cos {{{20}}^{\circ }}} \right)}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{16\left( {1-2\sin {{{30}}^{\circ }}\sin {{{10}}^{\circ }}} \right)}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{16\left( {\sin {{{90}}^{\circ }}-\sin {{{10}}^{\circ }}} \right)}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{32\cos {{{50}}^{\circ }}\sin {{{40}}^{\circ }}}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}$

$\displaystyle \ \ \ \ =\frac{{32{{{\sin }}^{2}}{{{40}}^{\circ }}}}{{{{{\sin }}^{2}}{{{40}}^{\circ }}}}\ \ \ \ \ \left[ {\because \cos {{{50}}^{\circ }}=\sin {{{40}}^{\circ }}} \right]$

$\displaystyle \ \ \ \ =32$

### Solving problem with compound angle formula

Find the value of
$\displaystyle (\ \text{i}\ )\ \ \ \ \frac{{\sin {{{80}}^{\circ }}}}{{\sin {{{20}}^{\circ }}}}-\frac{{\sqrt{3}}}{{2\sin {{{80}}^{\circ }}}}$

$\displaystyle (\ \text{ii}\ )\ \ \ \ \left( {\cot {{{25}}^{\circ }}-1} \right)\left( {\cot {{{24}}^{\circ }}-1} \right)\left( {\cot {{{20}}^{\circ }}-1} \right)\left( {\cot {{{21}}^{\circ }}-1} \right)$

Solution

$\displaystyle (\ \text{i}\ )\ \ \ \ \frac{{\sin {{{80}}^{\circ }}}}{{\sin {{{20}}^{\circ }}}}-\frac{{\sqrt{3}}}{{2\sin {{{80}}^{\circ }}}}$

$\displaystyle \ \ \ \ \ \ =\frac{{\cos {{{10}}^{\circ }}}}{{2\sin {{{10}}^{\circ }}\cos {{{10}}^{\circ }}}}-\frac{{\sqrt{3}}}{{2\cos {{{10}}^{\circ }}}}$

$\displaystyle \ \ \ \ \ \ =\frac{1}{{2\sin {{{10}}^{\circ }}}}-\frac{{\sqrt{3}}}{{2\cos {{{10}}^{\circ }}}}$

$\displaystyle \ \ \ \ \ \ =\frac{{\frac{1}{2}}}{{\sin {{{10}}^{\circ }}}}-\frac{{\frac{{\sqrt{3}}}{2}}}{{\cos {{{10}}^{\circ }}}}$

$\displaystyle \ \ \ \ \ \ =\frac{{\sin {{{30}}^{\circ }}}}{{\sin {{{10}}^{\circ }}}}-\frac{{\cos {{{30}}^{\circ }}}}{{\cos {{{10}}^{\circ }}}}$

$\displaystyle \ \ \ \ \ \ =\frac{{\sin {{{30}}^{\circ }}\cos {{{10}}^{\circ }}-\cos {{{30}}^{\circ }}\sin {{{10}}^{\circ }}}}{{\sin {{{10}}^{\circ }}\cos {{{10}}^{\circ }}}}$

$\displaystyle \ \ \ \ \ \ =\frac{{\sin \left( {{{{30}}^{\circ }}-{{{10}}^{\circ }}} \right)}}{{\sin {{{10}}^{\circ }}\cos {{{10}}^{\circ }}}}$

$\displaystyle \ \ \ \ \ \ =\frac{{\sin {{{20}}^{\circ }}}}{{\sin {{{10}}^{\circ }}\cos {{{10}}^{\circ }}}}$

$\displaystyle \ \ \ \ \ \ =\frac{{2\sin {{{10}}^{\circ }}\cos {{{10}}^{\circ }}}}{{\sin {{{10}}^{\circ }}\cos {{{10}}^{\circ }}}}$

$\displaystyle \ \ \ \ \ \ =2$

$\displaystyle (\ \text{ii}\ )\ \ \ \ \left( {\cot {{{25}}^{\circ }}-1} \right)\left( {\cot {{{24}}^{\circ }}-1} \right)\left( {\cot {{{20}}^{\circ }}-1} \right)\left( {\cot {{{21}}^{\circ }}-1} \right)$

$\displaystyle \ \ \ \ \ \ \ =\left( {\frac{1}{{\tan {{{25}}^{\circ }}}}-1} \right)\left( {\frac{1}{{\tan {{{24}}^{\circ }}}}-1} \right)\left( {\frac{1}{{\tan {{{20}}^{\circ }}}}-1} \right)\left( {\frac{1}{{\tan {{{21}}^{\circ }}}}-1} \right)$

$\displaystyle \ \ \ \ \ \ \ =\left( {\frac{{1-\tan {{{25}}^{\circ }}}}{{\tan {{{25}}^{\circ }}}}} \right)\left( {\frac{{1-\tan {{{24}}^{\circ }}}}{{\tan {{{24}}^{\circ }}}}} \right)\left( {\frac{{1-\tan {{{20}}^{\circ }}}}{{\tan {{{20}}^{\circ }}}}} \right)\left( {\frac{{1-\tan {{{21}}^{\circ }}}}{{\tan {{{21}}^{\circ }}}}} \right)$

$\displaystyle \ \ \ \ \ \ \ =\left( {\frac{{\left( {1-\tan {{{25}}^{\circ }}} \right)\left( {1-\tan {{{20}}^{\circ }}} \right)}}{{\tan {{{25}}^{\circ }}\tan {{{20}}^{\circ }}}}} \right)\left( {\frac{{\left( {1-\tan {{{24}}^{\circ }}} \right)\left( {1-\tan {{{21}}^{\circ }}} \right)}}{{\tan {{{24}}^{\circ }}\tan {{{21}}^{\circ }}}}} \right)$

$\displaystyle \ \ \ \ \ \ \ =\left( {\frac{{1-\tan {{{20}}^{\circ }}-\tan {{{25}}^{\circ }}+\tan {{{25}}^{\circ }}\tan {{{20}}^{\circ }}}}{{\tan {{{25}}^{\circ }}\tan {{{20}}^{\circ }}}}} \right)\left( {\frac{{1-\tan {{{21}}^{\circ }}-\tan {{{24}}^{\circ }}+\tan {{{24}}^{\circ }}\tan {{{21}}^{\circ }}}}{{\tan {{{24}}^{\circ }}\tan {{{21}}^{\circ }}}}} \right)$

$\displaystyle \ \ \ \ \ \ \ =\left( {\frac{{2\tan {{{25}}^{\circ }}\tan {{{20}}^{\circ }}}}{{\tan {{{25}}^{\circ }}\tan {{{20}}^{\circ }}}}} \right)\left( {\frac{{2\tan {{{24}}^{\circ }}\tan {{{21}}^{\circ }}}}{{\tan {{{24}}^{\circ }}\tan {{{21}}^{\circ }}}}} \right)$

$\displaystyle \ \ \ \ \ \ \ =4$

$\displaystyle *Note$

$\displaystyle \tan {{45}^{\circ }}=\tan \left( {{{{25}}^{\circ }}+{{{20}}^{\circ }}} \right)=\frac{{\tan {{{25}}^{\circ }}+\tan {{{20}}^{\circ }}}}{{1-\tan {{{25}}^{\circ }}\tan {{{20}}^{\circ }}}}$

$\displaystyle \tan {{45}^{\circ }}=\tan \left( {{{{24}}^{\circ }}+{{{21}}^{\circ }}} \right)=\frac{{\tan {{{24}}^{\circ }}+\tan {{{21}}^{\circ }}}}{{1-\tan {{{24}}^{\circ }}\tan {{{21}}^{\circ }}}}$

$\displaystyle \tan {{45}^{\circ }}=1$

$\displaystyle \therefore \ \ \tan {{25}^{\circ }}\tan {{20}^{\circ }}=1-\tan {{25}^{\circ }}-\tan {{20}^{\circ }}\ \ and\ \ \tan {{24}^{\circ }}\tan {{21}^{\circ }}=1-\tan {{24}^{\circ }}-\tan {{21}^{\circ }}\$