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Question: If \(\pi = {180^0}\) and \(A = \dfrac{\pi }{6}\), prove that \(\dfrac{{\left( {1 - \cos A} \right)\l...

If π=1800\pi = {180^0} and A=π6A = \dfrac{\pi }{6}, prove that (1−cos⁡A)(1+cos⁡A)(1−sin⁡A)(1+sin⁡A)=13\dfrac{{\left( {1 - \cos A} \right)\left( {1 + \cos A} \right)}}{{\left( {1 - \sin A} \right)\left( {1 + \sin A} \right)}} = \dfrac{1}{3}

Explanation

Solution

It is given in the question that π=1800\pi = {180^0} and A=π6A = \dfrac{\pi }{6}.
So, we have to prove that (1−cos⁡A)(1+cos⁡A)(1−sin⁡A)(1+sin⁡A)=13\dfrac{{\left( {1 - \cos A} \right)\left( {1 + \cos A} \right)}}{{\left( {1 - \sin A} \right)\left( {1 + \sin A} \right)}} = \dfrac{1}{3}.
we will take L.H.S and prove R.H.S.
Firstly, we will put the value of A=π6A = \dfrac{\pi }{6} in the equation. Then after, we will put the value of π=1800\pi = {180^0} in the equation which is formed by putting the value of A.
Finally, after solving the equation we will get the answer.

Formula Used:
(a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}

Complete step by step solution:
It is given in the question that π=1800\pi = {180^0} and A=π6A = \dfrac{\pi }{6}.
So, we have to prove that (1−cos⁡A)(1+cos⁡A)(1−sin⁡A)(1+sin⁡A)=13\dfrac{{\left( {1 - \cos A} \right)\left( {1 + \cos A} \right)}}{{\left( {1 - \sin A} \right)\left( {1 + \sin A} \right)}} = \dfrac{1}{3}.
Let, L.H.S,
Now, put the value of A=π6A = \dfrac{\pi }{6} in the above equation, we get
=(1−cos⁡π6)(1+cos⁡π6)(1−sin⁡π6)(1+sin⁡π6)= \dfrac{{\left( {1 - \cos \dfrac{\pi }{6}} \right)\left( {1 + \cos \dfrac{\pi }{6}} \right)}}{{\left( {1 - \sin \dfrac{\pi }{6}} \right)\left( {1 + \sin \dfrac{\pi }{6}} \right)}}
Now, put the value of π=1800\pi = {180^0} in the above equation, we get
=(1−cos⁡18006)(1+cos⁡18006)(1−sin⁡18006)(1+sin⁡18006)= \dfrac{{\left( {1 - \cos \dfrac{{{{180}^0}}}{6}} \right)\left( {1 + \cos \dfrac{{{{180}^0}}}{6}} \right)}}{{\left( {1 - \sin \dfrac{{{{180}^0}}}{6}} \right)\left( {1 + \sin \dfrac{{{{180}^0}}}{6}} \right)}}
=(1−cos⁡300)(1+cos⁡300)(1−sin⁡300)(1+sin⁡300)= \dfrac{{\left( {1 - \cos {{30}^0}} \right)\left( {1 + \cos {{30}^0}} \right)}}{{\left( {1 - \sin {{30}^0}} \right)\left( {1 + \sin {{30}^0}} \right)}}
Since, we know that cos⁡300=32\cos {30^0} = \dfrac{{\sqrt 3 }}{2} and sin⁡300=12\sin {30^0} = \dfrac{1}{2}. put this value in the above equation, we get
=(1−32)(1+32)(1−12)(1+12)= \dfrac{{\left( {1 - \dfrac{{\sqrt 3 }}{2}} \right)\left( {1 + \dfrac{{\sqrt 3 }}{2}} \right)}}{{\left( {1 - \dfrac{1}{2}} \right)\left( {1 + \dfrac{1}{2}} \right)}}
=(2−32)(2+32)(2−12)(2+12)= \dfrac{{\left( {\dfrac{{2 - \sqrt 3 }}{2}} \right)\left( {\dfrac{{2 + \sqrt 3 }}{2}} \right)}}{{\left( {\dfrac{{2 - 1}}{2}} \right)\left( {\dfrac{{2 + 1}}{2}} \right)}}
=(2−32)(2+32)(2−12)(2+12)= \dfrac{{\left( {\dfrac{{2 - \sqrt 3 }}{{{2}}}} \right)\left( {\dfrac{{2 + \sqrt 3 }}{{{2}}}} \right)}}{{\left( {\dfrac{{2 - 1}}{{{2}}}} \right)\left( {\dfrac{{2 + 1}}{{{2}}}} \right)}}
=(2−3)(2+3)(2−1)(2+1)= \dfrac{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}}{{\left( {2 - 1} \right)\left( {2 + 1} \right)}}
Now, using formula (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2} in the numerator, we get
=(2)2−(3)2(1)(3)= \dfrac{{{{\left( 2 \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}}}{{\left( 1 \right)\left( 3 \right)}}
=4−33= \dfrac{{4 - 3}}{3}
=13=R.H.S= \dfrac{1}{3} = R.H.S
⇒L.H.S=R.H.S\Rightarrow L.H.S = R.H.S

Note:
Some formula of cosine function:

  1. cos⁡(−x)=cos⁡x\cos \left( { - x} \right) = \cos x
  2. cos⁡(x+2kπ)=cos⁡x\cos \left( {x + 2k\pi } \right) = \cos x
  3. cos⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡y\cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y
  4. cos⁡(x−y)=cos⁡xcos⁡y+sin⁡xsin⁡y\cos \left( {x - y} \right) = \cos x\cos y + \sin x\sin y
  5. cos⁡2x=cos⁡2x−sin⁡2x=2cos⁡2x−1\cos 2x = {\cos ^2}x - {\sin ^2}x = 2{\cos ^2}x - 1
  6. cos⁡θ=1−t21+t2\cos \theta = \dfrac{{1 - {t^2}}}{{1 + {t^2}}}