Multiples And Sub-multiple Angles

Multiples And Sub-multiple Angles
 If A,2A, 3A ...... are called multiple angles, then \(\frac{A}{2},\,\frac{A}{3},\,\frac{A}{4}.....\) are called sub multiple angles.
    1) sin 2A = 2 sinA cosA =\(\frac{{2\tan A}}{{1 + {{\tan }^2}A}}\)
    2) cos 2A = cos²A - sin²A = 1- 2sin² A = 2cos² A - 1=\(\frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}\)
    3) tan 2A =\(\frac{{2\tan A}}{{1 - {{\tan }^2}A}}\)
    4) cos 2A =\(\frac{{{{\cot }^2}A - 1}}{{2\cot A}}\)
    5) sin 3A = 3 sinA - 4sin³A
    6) cos 3A = 4 cos³ A - 3 cosA
    7) tan3A =\(\frac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}\)
    8)cot 3A =\(\frac{{3\cot A - {{\cot }^3}A}}{{1 - 3{{\cot }^2}A}}\)
    9) \(\sin A = 2\sin \frac{A}{2}\,\,.\,\cos \frac{A}{2} = \frac{{2\tan \frac{A}{2}}}{{1 + {{\tan }^2}\frac{A}{2}}}\)
    10)cosA = \({\cos ^2}\frac{A}{2} - {\sin ^2}\frac{A}{2} = 1 - 2{\sin ^2}\frac{A}{2}\)
     \(= 2{\cos ^2}\frac{A}{2} - 1 = \frac{{1 - {{\tan }^2}\frac{A}{2}}}{{1 + {{\tan }^2}\frac{A}{2}}}\)          
    11)tanA=\(\frac{{2\tan \frac{A}{2}}}{{1 - {{\tan }^2}\frac{A}{2}}}\)
    12)cotA =\(\frac{{{{\cot }^2}\frac{A}{2} - 1}}{{2\cot \frac{A}{2}}}\)
    13) sinA= \(3\sin \frac{A}{3} - 4{\sin ^3}\frac{A}{3}\)
    14)cosA =\(4{\cos ^3}\frac{A}{3} - 3\,\cos \frac{A}{3}\)
    15) tanA = \(\frac{{3\tan \frac{A}{3} - {{\tan }^3}\frac{A}{3}}}{{1 - 3{{\tan }^2}\frac{A}{3}}}\)
    16) cotA= \(\frac{{3\cot \frac{A}{3} - {{\cot }^3}\frac{A}{3}}}{{1 - 3{{\cot }^2}\frac{A}{3}}}\)
Proof : We know,

 i) sin (A + B) = sinA . cosB + cosA . sinB
    Put A = B
    sin 2A = sinA cosA + cosA . sinA

                    sin2A = 2 sinA . cosA        ........(i)

ii)     We know
    cos (A + B) = cosA . cosB - sinA . sinB
    Put B = A
    cos2A = cosA . cosA - sinA. sinA = cos²A - sin²A
                    cos2A = cos² A - sin²A
                cos2A = cos²A - sin²A = 1- sin² A - sin²A                
                cos2A  = 1- 2 sin²
                 cos2A  = cos²A - sin²A
                           = cos²A - (1- cos²A)
               cos2A = 2cos²A -1
     cos2A = \(\frac{{{{\cos }^2}A - {{\sin }^2}A}}{1} = \frac{{\frac{{{{\cos }^2}A - {{\sin }^2}A}}{{{{\cos }^2}A}}}}{{\frac{1}{{{{\cos }^2}A}}}}\)
                    =\(\frac{{1 - {{\tan }^2}A}}{{{{\sec }^2}A}}\)
     \(\cos 2A = \frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}\)

    cos2A = cos²A - sin²A = 1- 2sin²A = 2cos²A -1=\(\frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}\)
     We know 
     tan(A+B) =\(\frac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\)
    put B = A,  

     \(\therefore \tan 2A = \frac{{\tan A + \tan A}}{{1 - \tan A.\tan A}} = \frac{{2\tan A}}{{1 - {{\tan }^2}A}}\)
     \(\tan 2A = \frac{{2\tan A}}{{1 - {{\tan }^2}A}} = \frac{{2\cot A}}{{{{\cot }^2}A - 1}}\,\,\,\,\,\,\,\,\,\,\, \to (iii)\)
   We know, 
    \(\cot (A + B) = \frac{{\cot A.\cot B - 1}}{{\cot B + \cot A}}\)
    put B = A
    \(\cot 2A = \frac{{\cot A.\cot A - 1}}{{\cot A + \cot A}} = \frac{{{{\cot }^2}A - 1}}{{2\cot A}} = \frac{{1 - {{\tan }^2}A}}{{2\tan A}}\)
    \(\therefore \,\,\,\,\,\,\cot 2A = \frac{{{{\cot }^2}A - 1}}{{2\cot A}} = \frac{{1 - {{\tan }^2}A}}{{2\tan A}}\,\,\,\,\,\,\,\, \to (iv)\)
   We know,
        \(\sin 3A = \sin (2A + A) = \sin 2A.\,\cos A + \cos 2A\,\,.\sin A\)
                        =  2 sin A cosA cosA + (1- 2sin²A) . sinA
                        = 2sin (1-sin²A)+ (1-2sin²A) . sinA
     \(\sin 3A = 3\sin A - 4{\sin ^3}A\,\,\,\,\,\,\,\,\,\,\,....(v)\)             

     We know, 

           cos3A   = cos(2A+A) = cos2A . cosA - sin2A. sinA
                      = (2cos²A - 1) cosA - 2sinA . cosA . sinA
                      = 2cos³A - cosA - 2(1-cos²A) . cosA
                      = 2 cos³A - cosA - 2cosA + 2cos³A
         \(\cos 3A = 4{\cos ^3}A - 3\cos A\,\,\,\,\,\,\, \to (vi)\)
     We know tan 3A = tan(2A + A)
                                \(= \frac{{\tan 2A + \tan A}}{{1 - \tan 2A.\,\tan A}}\)
                                \(= \frac{{\frac{{2\tan A}}{{1 - {{\tan }^2}A}} + \tan A}}{{1 - \left( {\frac{{2\tan A}}{{1 - {{\tan }^2}A}}} \right)\tan A}}\) 
                                \( = \frac{{\frac{{2\tan A + \tan (1 - {{\tan }^2}A)}}{{1 - {{\tan }^2}A}}}}{{\frac{{(1 - {{\tan }^2}A) - 2{{\tan }^2}A}}{{1 - {{\tan }^2}A}}}}\)
         \(\tan 3A = \frac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}\,\,\,\,\,\,\, \to \,\,(vii)\)   
   We know 
                 cos3A = cot(2A+A) = \(\frac{{\cot 2A.\cot A - 1}}{{\cot 2A + \cot A}}\)
                                     \(\frac{{\left( {\frac{{{{\cot }^2}A - 1}}{{2\cot A}}} \right).\cot A - 1}}{{\left( {\frac{{{{\cot }^2}A - 1}}{{2\cot A}}} \right) + \cot A}}\)
                                     \(\frac{{{{\cot }^3}A - 3\cot A}}{{3{{\cot }^2}A - 1}}\)
                                     \(\cot 3A = \frac{{3\cot A - {{\cot }^3}A}}{{1 - 3{{\cot }^2}A}}\,\,\,\,\,\,\,\,\,\, \to \,\,(viii)\)
       We know 
                 sin2A = 2 sinA cosA
                  put A =\(A \over 2\)
                  sin 2\(({A \over 2})\)=2sin \(A \over 2\).cos \(A \over 2\)
                 sinA = 2sin \(A \over 2\) . cos \(A \over 2\)
                \(\sin 2A = \frac{{2\tan A}}{{1 + {{\tan }^2}A}}\)
                put A = \(A \over 2\)
                     \(\sin \left( {2.\frac{A}{2}} \right) = \frac{{2\tan \frac{A}{2}}}{{1 + {{\tan }^2}\frac{A}{2}}}\)    
                    \(\sin A = \frac{{2\tan \frac{A}{2}}}{{1 + {{\tan }^2}\frac{A}{2}}}\)
    
                \(\sin A = 2\sin \frac{A}{2}\,\,\cos \frac{A}{2} = \frac{{2\tan \frac{A}{2}}}{{1 + {{\tan }^2}\frac{A}{2}}}\,\,\,\,\,\, \to (ix)\)

We know 
               \(\cos 2A = {\cos ^2}A - {\sin ^2}A = 1 - 2{\sin ^2}A = 2{\cos ^2}A - 1 = \frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}\)
               Put A=\(A \over 2\)
               \(cos A = {\cos ^2}\frac{A}{2} - {\sin ^2}\frac{A}{2} = 1 - 2{\sin ^2}\frac{A}{2} = 2{\cos ^2}\frac{A}{2} - 1 = \frac{{1 - {{\tan }^2}\frac{A}{2}}}{{1 + {{\tan }^2}\frac{A}{2}}}\,\, \to (x)\)

we know 
               tan 2A=    \(\frac{{2\tan A}}{{1 - {{\tan }^2}A}}\)
                put A=   \(A \over 2\) 

                 \(\tan A = \frac{{2\tan \frac{A}{2}}}{{1 - {{\tan }^2}\frac{A}{2}}}\,\,\,\, \to (xi)\)
We know
               \(\cot 2A = \frac{{{{\cot }^2}A - 1}}{{2\cot A}}\)
             put A=\(A\over 2\)     

             \(\cot A = \frac{{{{\cot }^2}\frac{A}{2} - 1}}{{2\cot \frac{A}{2}}}\,\,\,\, \to (xii)\)
We know 
              sin3A=3 sinA - 4 sin³A
               Put  A = \(A \over 3\)

              sinA = 3sin \(A\over 3\) - 4sin³ \(A\over 3\) -----  (xiii)

We know     
              cos3A = 4 cos³A - 3cosA
             put A =\(A \over 3\) 

             cosA =4 cos^{3\(A \over 3\)} - 3 cos\(A\over 3\) ------    (xiv)

We know, tan 3A = \(\frac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}\)
                 put A=\(A\over 3\)
                 \(\tan A = \frac{{3\tan \frac{A}{3} - {{\tan }^3}\frac{A}{3}}}{{1 - 3{{\tan }^2}\frac{A}{3}}}\,\,\,\,\, \to (xv)\)
we know, \(\cot 3A = \frac{{3\cot A - {{\cot }^3}A}}{{1 - 3{{\cot }^2}A}}\)
             put   A = \(A\over 3\)

\(\cot A = \frac{{3\cot \frac{A}{3} - {{\cot }^3}\frac{A}{3}}}{{1 - 3{{\cot }^2}\frac{A}{3}}}\,\,\,\,\, \to (xvi)\)

Theorem : When cos2A is known, then the value of sinA, cosA tan A and cotA in terms of cos2A

i)   \(sin A = \pm \sqrt {\frac{{1 - \cos 2A}}{2}} \)  
    We know,
    cos2A = 1- 2 sin²A
    \(\Rightarrow 2{\sin ^2}A = 1 - \cos 2A\)
    \(\begin{gathered} \Rightarrow 2{\sin ^2}A = 1 - \cos 2A \hfill \\ \Rightarrow {\sin ^2}A = \frac{{1 - \cos 2A}}{2} \hfill \\ \end{gathered} \)
    \( \Rightarrow \,\,\,\sin A = \pm \sqrt {\frac{{1 - \cos 2A}}{2}} \)

ii)\(\cos A = \pm \sqrt {\frac{{1 + \cos 2A}}{2}} \)    

We know
    cos2A = 2cos²A - 1
     \(\Rightarrow \,\,\,\,2{\cos ^2}A = 1 + \cos 2A\)
       \(\Rightarrow \,\,\,\,{\cos ^2}A = \frac{{1 + \cos 2A}}{2}\)
       \(\Rightarrow \,\,\,\,\cos A = \pm \sqrt {\frac{{1 + \cos 2A}}{2}} \)
iii)\(\tan A = \pm \sqrt {\frac{{1 - \cos 2A}}{{1 + \cos 2A}}} \)    
      we know \(\tan A = \frac{{\sin A}}{{\cos A}}\)            

     \(= \pm \frac{{\sqrt {\frac{{1 - \cos 2A}}{2}} }}{{\sqrt {\frac{{1 + \cos 2A}}{2}} }}\)       

      \(\tan A = \pm \sqrt {\frac{{1 - \cos 2A}}{{1 + \cos 2A}}} \)          

 iv) We know
\(% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4yaiaac+ % gacaGG0bGaamyqaiabg2da9maalaaabaGaaGymaaqaaiGacshacaGG % HbGaaiOBaiaadgeaaaaaaa!3EF6! \[\cot A = \frac{1}{{\tan A}}\)
   \(\cot A = \pm \sqrt {\frac{{1 + \cos 2A}}{{1 - \cos 2A}}} \) 
Theorem :      
Prove that i) \(\sin {18^0} = \frac{{\sqrt 5 - 1}}{4}\)    ii)  \(\cos {18^0} = \frac{{\sqrt {10 + 2\sqrt 5 } }}{4}\)      iii)\(\cos {36^0} = \frac{{\sqrt 5 + 1}}{4}\)
    iv) \(\sin {36^0} = \frac{{\sqrt {10 - 2\sqrt 5 } }}{4}\)
Proof : If A=18⁰
     5A = 90⁰
     3A + 2A =90⁰
     sin2A= sin(90⁰-3A) =cos3A
     2sinA cosA = 4 cos³A - 3cosA
     2sinA = 4cos² A-3                  [divide by cosA on both sides]
     2 sinA = 4 (1-sin²) -3
     4 sin² A + 2 sinA - 1 =0
     a = 4, b = 2, c = -1  
     \(\therefore \,\sin A\,\, = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\)

       \(\sin A = \frac{{ - 2 \pm \sqrt {4 + 16} }}{{2.1}}\)
     \(= \frac{{ - 2 \pm \sqrt {20} }}{{2 \times 4}}\) \( = \frac{{ - 2 \pm 2\sqrt 5 }}{8}\) \(= \frac{{ - 1 \pm \sqrt 5 }}{4}\) \(\frac{{\sqrt 5 - 1}}{4}(or) - \frac{{ - 1 - \sqrt 5 }}{4}\)
      18⁰ belongs to Q₁ . sin18⁰ is positive 
       \(\therefore \,\,\,\sin {18^0} = \frac{{\sqrt 5 - 1}}{4}\)
ii)    sin18⁰ = \(\frac{{\sqrt 5 - 1}}{4}\)
          cos18⁰ =\(\sqrt {1 - {{\sin }^2}{{18}^0}} \)
            =\(\sqrt {1 - {{\left( {\frac{{\sqrt 5 - 1}}{4}} \right)}^2}} \)
            =\(\sqrt {1 - \left( {\frac{{6 - 2\sqrt 5 }}{{16}}} \right)} \)
            =\(\sqrt {\frac{{10 + 2\sqrt 5 }}{{16}}} \)
    cos18⁰ = \(\sqrt {\frac{{10 + 2\sqrt 5 }}{{16}}} \)
    \(\cos {18^0} = \frac{{\sqrt {10 + 2\sqrt 5 } }}{4}\)
 iii)    cos36⁰ = cos2(18⁰)
            = 1- 2sin²18⁰
            = \(1 - 2{\left( {\frac{{\sqrt 5 - 1}}{4}} \right)^2}\)
            = \(1 - 2\left( {\frac{{6 - 2\sqrt 5 }}{{16}}} \right)\)
            =\(1 - \left( {\frac{{6 - 2\sqrt 5 }}{8}} \right)\)
            \(\frac{{8 - 6 + 2\sqrt 5 }}{8}\)
    cos36⁰ =\(\frac{{\sqrt 5 + 1}}{4}\)
(iv)\(\sin {36^0} = \sqrt {1 - {{\cos }^2}{{36}^0}} \)  
\(\sqrt {1 - {{\left( {\frac{{\sqrt 5 + 1}}{4}} \right)}^2}} \)
           \(\sqrt {1 - \left( {\frac{{6 + 2\sqrt 5 }}{{16}}} \right)} \)
           \(\sqrt {\frac{{10 - 2\sqrt 5 }}{{16}}} \)
     sin36^{0 \(=\frac{{\sqrt {10 - 2\sqrt 5 } }}{4}\)    
Theorem :
Prove that : i) \(\sin 22{\frac{1}{2}^0} = \sqrt {\frac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} \)    ii)  \(\cos 22{\frac{1}{2}^0} = \sqrt {\frac{{\sqrt 2 + 1}}{{2\sqrt 2 }}} \)  iii) \(\tan 22{\frac{1}{2}^0} = \sqrt 2 - 1\)
iv) \(\cot 22{\frac{1}{2}^0} = \sqrt 2 + 1\)}
Proof : We know, 
\([\sin \frac{A}{2} = \pm \sqrt {\frac{{1 - \cos A}}{2}} \)
Put A= 45^{0
\(\sin 22{\frac{1}{2}^0} = \sqrt {\frac{{1 - \cos {{45}^0}}}{2}} \)
\(\sqrt {\frac{{1 - \frac{1}{{\sqrt 2 }}}}{2}} \)}
\(\sin 22{\frac{1}{2}^0} = \sqrt {\frac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} \)
ii)\(\cos \frac{A}{2} = \sqrt {\frac{{1 + \cos A}}{2}} \)

A=45^{0
\(\cos 22{\frac{1}{2}^0} = \sqrt {\frac{{1 + \cos {{45}^0}}}{2}} \)}

=\(\sqrt {\frac{{1 + \frac{1}{{\sqrt 2 }}}}{2}} \)
\(\cos 22{\frac{1}{2}^0} = \sqrt {\frac{{\sqrt 2 + 1}}{{2\sqrt 2 }}} \)
iii)\(\tan \frac{A}{2} = \sqrt {\frac{{1 - \cos A}}{{1 + \cos A}}} \)
A=45^{0
\(\tan 22{\frac{1}{2}^0} = \sqrt {\frac{{1 - \cos {{45}^0}}}{{1 + \cos {{45}^0}}}} \)}
\(=\sqrt {\frac{{1 - \frac{1}{{\sqrt 2 }}}}{{1 + \frac{1}{{\sqrt 2 }}}}} \) =\(\sqrt {\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \)
=\(\sqrt {\frac{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}}} \) =\(\sqrt {\frac{{{{\left( {\sqrt 2 - 1} \right)}^2}}}{{{{\sqrt 2 }^2} - {1^2}}}} \)
\(\tan 22{\frac{1}{2}^0} = \sqrt 2 - 1\)
Note:\(\cot 22{\frac{1}{2}^0} = \frac{1}{{\tan 22{{\frac{1}{2}}^0}}} = \frac{1}{{\sqrt 2 - 1}}\)
\(= \frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}}\)
\(\cot 22{\frac{1}{2}^0} = \sqrt 2 + 1\)
Note:\(\sin 22{\frac{1}{2}^0} = \cos 67{\frac{1}{2}^0}\)
\(= \frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}}\)
\(\cot 22{\frac{1}{2}^0} = \sqrt 2 + 1\)
Note:\(\sin 22{\frac{1}{2}^0} = \cos 67{\frac{1}{2}^0}\)
\(\cos 22{\frac{1}{2}^0} = \sin 67{\frac{1}{2}^0}\)
\(\tan 22{\frac{1}{2}^0} = \cot 67{\frac{1}{2}^0}\)
\(\cot 22{\frac{1}{2}^0} = \tan 67{\frac{1}{2}^0}\)
Theorem 
 Prove that :i) \(\sin 7{\frac{1}{2}^0} = \frac{{\sqrt {4 - \sqrt 6 - \sqrt 2 } }}{{2\sqrt 2 }}\)    ii)\(\cos 7{\frac{1}{2}^0} = \frac{{\sqrt {4 + \sqrt 6 + \sqrt 2 } }}{{2\sqrt 2 }}\)
    iii)\(\tan 7{\frac{1}{2}^0} = \sqrt 2 - \sqrt 3 - \sqrt 4 + \sqrt 6 = \left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 2 - 1} \right)\)
    iv)\(\cot 7{\frac{1}{2}^0} = \sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 = \left( {\sqrt 3 + \sqrt 2 } \right)\left( {\sqrt 2 + 1} \right)\)
Proof : We have
\(\cos {15^0} = \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}\)
we Know
\(\sin \frac{A}{2} = \pm \sqrt {\frac{{1 - \cos A}}{2}} \)

\(\sin 7{\frac{1}{2}^0} = \sqrt {\frac{{1 - \cos {{15}^0}}}{2}} \)

\( = \, \pm \sqrt {\frac{{1 - \left( {\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)}}{2}} \)
\( = \, \pm \sqrt {\frac{{2\sqrt 2 - \sqrt 3 - 1}}{{4\sqrt 2 }}} \)
\(\sin 7{\frac{1}{2}^0} = \frac{{\sqrt {4 - \sqrt 6 - \sqrt 2 } }}{{2\sqrt 2 }}\)
ii)we have, 
\(\cos \frac{A}{2} = \pm \sqrt {\frac{{1 + \cos A}}{2}} \)
putting A =15^{0 
\(\cos 7\frac{1}{2}\,\,\, = \,\,\,\,\sqrt {\frac{{1 + \left( {\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)}}{2}} \)}
\(\cos 7\frac{1}{2} = \sqrt {\frac{{2\sqrt 2 + \sqrt 3 + 1}}{{4\sqrt 2 }}} \)
\(= \sqrt {\frac{{4 + \sqrt 6 + \sqrt 2 }}{8}} \)
\(\cos 7{\frac{1}{2}^0} = \frac{{\sqrt {4 + \sqrt 6 + 2\sqrt 2 } }}{{2\sqrt 2 }}\)
iii)\(\tan 7{\frac{1}{2}^0} = \frac{{\sin 7{{\frac{1}{2}}^0}}}{{\cos 7{{\frac{1}{2}}^0}}}\)
=\(\frac{{2{{\sin }^2}7{{\frac{1}{2}}^0}}}{{2\sin 7{{\frac{1}{2}}^0}\,.\,\cos 7{{\frac{1}{2}}^0}}}\)=\(\frac{{1 - \cos {{15}^0}}}{{\sin {{15}^0}}}\)
\(\frac{{1 - \left( {\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)}}{{\left( {\frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)}}\)
=\(\frac{{2\sqrt 2 - \sqrt 3 - 1}}{{\sqrt 3 - 1}}\)
\(=\frac{{\left( {2\sqrt 2 - \sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}\)
=\(\frac{{2\sqrt 6 + 2\sqrt 2 - 3 - \sqrt 3 - \sqrt 3 - 1}}{2}\)
=\(\frac{{2\sqrt 6 - 2\sqrt 3 + 2\sqrt 2 - 4}}{2}\)
=\(\frac{{2\left( {\sqrt 6 - \sqrt 3 + \sqrt 2 - 2} \right)}}{2}\)
=\(\sqrt 6 - \sqrt 3 + \sqrt 2 - 2\)
=\(\tan 7{\frac{1}{2}^0} = \left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 2 - 1} \right)\)
(iv)\(\cot 7{\frac{1}{2}^0} = \frac{{\cos 7{{\frac{1}{2}}^0}}}{{\sin 7{{\frac{1}{2}}^0}}}\)
=\(\frac{{2{{\cos }^2}7{{\frac{1}{2}}^0}}}{{2\sin 7{{\frac{1}{2}}^0}.\,\cos 7{{\frac{1}{2}}^0}}}\)
=\(\frac{{1 + \cos {{15}^0}}}{{\sin {{15}^0}}}\)=\(\frac{{1 + \left( {\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)}}{{\left( {\frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)}}\)=\(\frac{{\left( {2\sqrt 2 + \sqrt 3 + 1} \right)\left( {\sqrt 3 + 1} \right)}}{{3 - 1}}\)
=\(\frac{{2\sqrt 6 + 2\sqrt 2 + 2\sqrt 3 + 4}}{2}\)=\(\sqrt 6 + \sqrt 2 + \sqrt 3 + \sqrt 4 \)
=\(\cot 7{\frac{1}{2}^0} = \left( {\sqrt 3 + \sqrt 2 } \right)\left( {\sqrt 2 + 1} \right)\)
Note : i) \(\sin 7{\frac{1}{2}^0} = \cos 82{\frac{1}{2}^0}\)
    ii)\(\cos 7{\frac{1}{2}^0} = \sin 82{\frac{1}{2}^0}\)
    iii) \(\tan 7{\frac{1}{2}^0} = \cot 82{\frac{1}{2}^0}\)
    iv)\(\cot 7{\frac{1}{2}^0} = \tan 82{\frac{1}{2}^0}\)