Respuesta :
[tex]{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}[/tex]
[tex] \star \: \tt \cot \theta = \dfrac{7}{8} [/tex]
[tex] {\large{\textsf{\textbf{\underline{\underline{To \: Evaluate :}}}}}}[/tex]
[tex] \star \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) }[/tex]
[tex]{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}[/tex]
Consider a [tex]\triangle[/tex] ABC right angled at C and [tex]\sf \angle \: B = \theta [/tex]
Then,
‣ Base [B] = BC
‣ Perpendicular [P] = AC
‣ Hypotenuse [H] = AB
[tex] \therefore \tt \cot \theta = \dfrac{Base}{ Perpendicular} = \dfrac{BC}{AC} = \dfrac{7}{8}[/tex]
Let,
Base = 7k and Perpendicular = 8k, where k is any positive integer
In [tex]\triangle[/tex] ABC, H² = B² + P² by Pythagoras theorem
[tex] \longrightarrow \tt {AB}^{2} = {BC}^{2} + {AC}^{2} [/tex]
[tex] \longrightarrow \tt {AB}^{2} = {(7k)}^{2} + {(8k)}^{2} [/tex]
[tex]\longrightarrow \tt {AB}^{2} = 49{k}^{2} + 64{k}^{2} [/tex]
[tex]\longrightarrow \tt {AB}^{2} = 113{k}^{2} [/tex]
[tex]\longrightarrow \tt AB = \sqrt{113 {k}^{2} } [/tex]
[tex]\longrightarrow \tt AB = \red{ \sqrt{113} \: k}[/tex]
Calculating Sin [tex]\sf \theta [/tex]
[tex] \longrightarrow \tt \sin \theta = \dfrac{Perpendicular}{Hypotenuse}[/tex]
[tex] \longrightarrow \tt \sin \theta = \dfrac{AC}{AB}[/tex]
[tex]\longrightarrow \tt \sin \theta = \dfrac{8 \cancel{k}}{ \sqrt{113} \: \cancel{ k } }[/tex]
[tex]\longrightarrow \tt \sin \theta = \purple{ \dfrac{8}{ \sqrt{113} } }[/tex]
Calculating Cos [tex]\sf \theta [/tex]
[tex] \longrightarrow \tt \cos \theta = \dfrac{Base}{Hypotenuse}[/tex]
[tex] \longrightarrow \tt \cos \theta = \dfrac{BC}{ AB} [/tex]
[tex] \longrightarrow \tt \cos \theta = \dfrac{7 \cancel{k}}{ \sqrt{113} \: \cancel{k } }[/tex]
[tex]\longrightarrow \tt \cos \theta = \purple{ \dfrac{7}{ \sqrt{113} } }[/tex]
Solving the given expression :-
[tex] \longrightarrow \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) } [/tex]
Putting,
• Sin [tex]\sf \theta [/tex] = [tex]\dfrac{8}{ \sqrt{113} }[/tex]
• Cos [tex]\sf \theta [/tex] = [tex]\dfrac{7}{ \sqrt{113} }[/tex]
[tex] \longrightarrow \: \tt \dfrac{ \bigg(1 + \dfrac{8}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{8}{ \sqrt{133}} \bigg) }{\bigg(1 + \dfrac{7}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{7}{ \sqrt{133}} \bigg)} [/tex]
Using (a + b ) (a - b ) = a² - b²
[tex]\longrightarrow \: \tt \dfrac{ { \bigg(1 \bigg)}^{2} - { \bigg( \dfrac{8}{ \sqrt{133} } \bigg)}^{2} }{ { \bigg(1 \bigg)}^{2} - { \bigg( \dfrac{7}{ \sqrt{133} } \bigg)}^{2} } [/tex]
[tex]\longrightarrow \: \tt \dfrac{1 - \dfrac{64}{113} }{ 1 - \dfrac{49}{113} } [/tex]
[tex]\longrightarrow \: \tt \dfrac{ \dfrac{113 - 64}{113} }{ \dfrac{113 - 49}{113} } [/tex]
[tex]\longrightarrow \: \tt { \dfrac { \dfrac{49}{113} }{ \dfrac{64}{113} } }[/tex]
[tex]\longrightarrow \: \tt { \dfrac{49}{113} }÷{ \dfrac{64}{113} }[/tex]
[tex]\longrightarrow \: \tt \dfrac{49}{ \cancel{113}} \times \dfrac{ \cancel{113}}{64} [/tex]
[tex]\longrightarrow \: \tt \dfrac{49}{64} [/tex]
[tex]\qquad \: \therefore \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) } = \pink{\dfrac{49}{64} }[/tex]
[tex]\begin{gathered} {\underline{\rule{300pt}{4pt}}} \end{gathered} [/tex]
[tex] {\large{\textsf{\textbf{\underline{\underline{We \: know :}}}}}}[/tex]
✧ Basic Formulas of Trigonometry is given by :-
[tex]\begin{gathered}\begin{gathered}\boxed { \begin{array}{c c} \\ \bigstar \: \sf{ In \:a \:Right \:Angled \: Triangle :} \\ \\ \sf {\star Sin \theta = \dfrac{Perpendicular}{Hypotenuse}} \\\\ \sf{ \star \cos \theta = \dfrac{ Base }{Hypotenuse}}\\\\ \sf{\star \tan \theta = \dfrac{Perpendicular}{Base}}\\\\ \sf{\star \cosec \theta = \dfrac{Hypotenuse}{Perpendicular}} \\\\ \sf{\star \sec \theta = \dfrac{Hypotenuse}{Base}}\\\\ \sf{\star \cot \theta = \dfrac{Base}{Perpendicular}} \end{array}}\\\end{gathered} \end{gathered}[/tex]
[tex]{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}[/tex]
✧ Figure in attachment
[tex]\begin{gathered} {\underline{\rule{200pt}{1pt}}} \end{gathered} [/tex]
