# Show that:

$\cot \left( {A + {{15}^0}} \right) - \tan \left( {A - {{15}^0}} \right) = \dfrac{{4\cos 2A}}{{1 + 2\sin 2A}}$

Answer

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Hint – In this question apply some basic properties of trigonometric identities such as $2\cos C\cos D = \cos \left( {C + D} \right) + \cos \left( {C - D} \right),$$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }},{\text{ }}\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ to reach the solution of the question.

Given equation is

$\cot \left( {A + {{15}^0}} \right) - \tan \left( {A - {{15}^0}} \right) = \dfrac{{4\cos 2A}}{{1 + \sin 2A}}$

Consider L.H.S

$ = \cot \left( {A + {{15}^0}} \right) - \tan \left( {A - {{15}^0}} \right)$

As we know $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }},{\text{ }}\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ , so apply these properties in the above equation we have,

\[ = \dfrac{{\cos \left( {A + {{15}^0}} \right)}}{{\sin \left( {A + {{15}^0}} \right)}} - \dfrac{{\sin \left( {A - {{15}^0}} \right)}}{{\cos \left( {A - {{15}^0}} \right)}} = \dfrac{{\cos \left( {A + {{15}^0}} \right)\cos \left( {A - {{15}^0}} \right) - \sin \left( {A + {{15}^0}} \right)\sin \left( {A - {{15}^0}} \right)}}{{\cos \left( {A - {{15}^0}} \right)\sin \left( {A + {{15}^0}} \right)}}\]

Now multiply and divide by 2 in above equation we have

\[ = \dfrac{{2\cos \left( {A + {{15}^0}} \right)\cos \left( {A - {{15}^0}} \right) - 2\sin \left( {A + {{15}^0}} \right)\sin \left( {A - {{15}^0}} \right)}}{{2\cos \left( {A - {{15}^0}} \right)\sin \left( {A + {{15}^0}} \right)}}\]

Now we all know that

$

2\cos C\cos D = \cos \left( {C + D} \right) + \cos \left( {C - D} \right), \\

2\sin C\sin D = \cos \left( {C - D} \right) - \cos \left( {C + D} \right){\text{ and}} \\

2\cos C\sin D = \sin \left( {C + D} \right) - \sin \left( {C - D} \right) \\

$

So, apply these properties in above equation we have,

\[ = \dfrac{{\cos \left( {2A} \right) + \cos \left( {{{30}^0}} \right) - \left( {\cos \left( {{{30}^0}} \right) - \cos \left( {2{A^0}} \right)} \right)}}{{\sin \left( {2A} \right) - \sin \left( { - {{30}^0}} \right)}}\]

Now as we know that $\sin \left( { - \theta } \right) = - \sin \theta $ , so apply this property in the above equation we have,

$ = \dfrac{{2\cos 2A}}{{\sin \left( {2A} \right) + \sin \left( {{{30}^0}} \right)}}$

Now we all know that $\sin {30^0} = \dfrac{1}{2}$

$ = \cot \left( {A + {{15}^0}} \right) - \tan \left( {A - {{15}^0}} \right) = = \dfrac{{2\cos 2A}}{{\sin \left( {2A} \right) + \dfrac{1}{2}}} = \dfrac{{4\cos 2A}}{{2\sin 2A + 1}}$

= R.H.S

Hence Proved

Note – In such types of questions the key concept we have to remember is that always recall all the properties of trigonometric identities which is all stated above then using these properties simplify the L.H.S part of the given equation we will get the required R.H.S part of the equation.

Given equation is

$\cot \left( {A + {{15}^0}} \right) - \tan \left( {A - {{15}^0}} \right) = \dfrac{{4\cos 2A}}{{1 + \sin 2A}}$

Consider L.H.S

$ = \cot \left( {A + {{15}^0}} \right) - \tan \left( {A - {{15}^0}} \right)$

As we know $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }},{\text{ }}\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ , so apply these properties in the above equation we have,

\[ = \dfrac{{\cos \left( {A + {{15}^0}} \right)}}{{\sin \left( {A + {{15}^0}} \right)}} - \dfrac{{\sin \left( {A - {{15}^0}} \right)}}{{\cos \left( {A - {{15}^0}} \right)}} = \dfrac{{\cos \left( {A + {{15}^0}} \right)\cos \left( {A - {{15}^0}} \right) - \sin \left( {A + {{15}^0}} \right)\sin \left( {A - {{15}^0}} \right)}}{{\cos \left( {A - {{15}^0}} \right)\sin \left( {A + {{15}^0}} \right)}}\]

Now multiply and divide by 2 in above equation we have

\[ = \dfrac{{2\cos \left( {A + {{15}^0}} \right)\cos \left( {A - {{15}^0}} \right) - 2\sin \left( {A + {{15}^0}} \right)\sin \left( {A - {{15}^0}} \right)}}{{2\cos \left( {A - {{15}^0}} \right)\sin \left( {A + {{15}^0}} \right)}}\]

Now we all know that

$

2\cos C\cos D = \cos \left( {C + D} \right) + \cos \left( {C - D} \right), \\

2\sin C\sin D = \cos \left( {C - D} \right) - \cos \left( {C + D} \right){\text{ and}} \\

2\cos C\sin D = \sin \left( {C + D} \right) - \sin \left( {C - D} \right) \\

$

So, apply these properties in above equation we have,

\[ = \dfrac{{\cos \left( {2A} \right) + \cos \left( {{{30}^0}} \right) - \left( {\cos \left( {{{30}^0}} \right) - \cos \left( {2{A^0}} \right)} \right)}}{{\sin \left( {2A} \right) - \sin \left( { - {{30}^0}} \right)}}\]

Now as we know that $\sin \left( { - \theta } \right) = - \sin \theta $ , so apply this property in the above equation we have,

$ = \dfrac{{2\cos 2A}}{{\sin \left( {2A} \right) + \sin \left( {{{30}^0}} \right)}}$

Now we all know that $\sin {30^0} = \dfrac{1}{2}$

$ = \cot \left( {A + {{15}^0}} \right) - \tan \left( {A - {{15}^0}} \right) = = \dfrac{{2\cos 2A}}{{\sin \left( {2A} \right) + \dfrac{1}{2}}} = \dfrac{{4\cos 2A}}{{2\sin 2A + 1}}$

= R.H.S

Hence Proved

Note – In such types of questions the key concept we have to remember is that always recall all the properties of trigonometric identities which is all stated above then using these properties simplify the L.H.S part of the given equation we will get the required R.H.S part of the equation.

Last updated date: 03rd Oct 2023

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