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Question: Show that: \(\cot \left( {A + {{15}^0}} \right) - \tan \left( {A - {{15}^0}} \right) = \dfrac{{4\c...

Show that:
cot⁡(A+150)−tan⁡(A−150)=4cos⁡2A1+2sin⁡2A\cot \left( {A + {{15}^0}} \right) - \tan \left( {A - {{15}^0}} \right) = \dfrac{{4\cos 2A}}{{1 + 2\sin 2A}}

Explanation

Solution

Hint – In this question apply some basic properties of trigonometric identities such as 2cos⁡Ccos⁡D=cos⁡(C+D)+cos⁡(C−D),2\cos C\cos D = \cos \left( {C + D} \right) + \cos \left( {C - D} \right), cot⁡θ=cos⁡θsin⁡θ, tan⁡θ=sin⁡θcos⁡θ\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⁡(A+150)−tan⁡(A−150)=4cos⁡2A1+sin⁡2A\cot \left( {A + {{15}^0}} \right) - \tan \left( {A - {{15}^0}} \right) = \dfrac{{4\cos 2A}}{{1 + \sin 2A}}
Consider L.H.S
=cot⁡(A+150)−tan⁡(A−150)= \cot \left( {A + {{15}^0}} \right) - \tan \left( {A - {{15}^0}} \right)
As we know cot⁡θ=cos⁡θsin⁡θ, tan⁡θ=sin⁡θcos⁡θ\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,
=cos⁡(A+150)sin⁡(A+150)−sin⁡(A−150)cos⁡(A−150)=cos⁡(A+150)cos⁡(A−150)−sin⁡(A+150)sin⁡(A−150)cos⁡(A−150)sin⁡(A+150)= \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
=2cos⁡(A+150)cos⁡(A−150)−2sin⁡(A+150)sin⁡(A−150)2cos⁡(A−150)sin⁡(A+150)= \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
2cos⁡Ccos⁡D=cos⁡(C+D)+cos⁡(C−D), 2sin⁡Csin⁡D=cos⁡(C−D)−cos⁡(C+D) and 2cos⁡Csin⁡D=sin⁡(C+D)−sin⁡(C−D)  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,
=cos⁡(2A)+cos⁡(300)−(cos⁡(300)−cos⁡(2A0))sin⁡(2A)−sin⁡(−300)= \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⁡(−θ)=−sin⁡θ\sin \left( { - \theta } \right) = - \sin \theta , so apply this property in the above equation we have,
=2cos⁡2Asin⁡(2A)+sin⁡(300)= \dfrac{{2\cos 2A}}{{\sin \left( {2A} \right) + \sin \left( {{{30}^0}} \right)}}
Now we all know that sin⁡300=12\sin {30^0} = \dfrac{1}{2}
=cot⁡(A+150)−tan⁡(A−150)==2cos⁡2Asin⁡(2A)+12=4cos⁡2A2sin⁡2A+1= \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.