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Question: Prove that: \(\dfrac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\left( \sin A\cos A \right)-\left( \sin B\cos B ...

Prove that: sin⁡2A−sin⁡2B(sin⁡Acos⁡A)−(sin⁡Bcos⁡B)=tan⁡(A+B)\dfrac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\left( \sin A\cos A \right)-\left( \sin B\cos B \right)}=\tan \left( A+B \right)

Explanation

Solution

Here we have to prove the given trigonometric equation. We will take the LHSLHS value and simplify it using the double angles formula. Firstly we will convert the numerator into cosine term and denominator into sine term using the double angle formula. Then we will use the formula for converting the sum value into product value in the numerator and denominator. Finally by using the relation between sine, cosine and tangent function we will get our desired answer.

Complete step by step answer:
We have to prove the equation given below:
sin⁡2A−sin⁡2B(sin⁡Acos⁡A)−(sin⁡Bcos⁡B)=tan⁡(A+B)\dfrac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\left( \sin A\cos A \right)-\left( \sin B\cos B \right)}=\tan \left( A+B \right)
Taking the LHSLHS value we get,
sin⁡2A−sin⁡2B(sin⁡Acos⁡A)−(sin⁡Bcos⁡B)\dfrac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\left( \sin A\cos A \right)-\left( \sin B\cos B \right)}…..(1)\left( 1 \right)
The double angle formulas are given below:
sin⁡2θ=12(1−cos⁡2θ){{\sin }^{2}}\theta =\dfrac{1}{2}\left( 1-\cos 2\theta \right)
Using the above formula in equation (1) we get,
=12(1−cos⁡2A)−12(1−cos⁡2B)sin⁡Acos⁡A−sin⁡Bcos⁡B= \dfrac{\dfrac{1}{2}\left( 1-\cos 2A \right)-\dfrac{1}{2}\left( 1-\cos 2B \right)}{\sin A\cos A-\sin B\cos B}
Next we will make the denominator value in terms of sine function by using the formula 2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta =\sin 2\theta .
So multiply and divide the each term of the denominator by 22 as follows:
=1−cos⁡2A−1+cos⁡2B22sin⁡Acos⁡A2−2sin⁡Bcos⁡B2= \dfrac{\dfrac{1-\cos 2A-1+\cos 2B}{2}}{\dfrac{2\sin A\cos A}{2}-\dfrac{2\sin B\cos B}{2}}
Using the formula 2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta =\sin 2\theta we get,
=1−cos⁡2A−1+cos⁡2B2sin⁡2A2−sin⁡2B2= \dfrac{\dfrac{1-\cos 2A-1+\cos 2B}{2}}{\dfrac{\sin 2A}{2}-\dfrac{\sin 2B}{2}}
=−cos⁡2A+cos⁡2B2sin⁡2A−sin⁡2B2= \dfrac{\dfrac{-\cos 2A+\cos 2B}{2}}{\dfrac{\sin 2A-\sin 2B}{2}}
Cancelling out 22 we get,
=−(cos⁡2A−cos⁡2B)sin⁡2A−sin⁡2B= \dfrac{-\left( \cos 2A-\cos 2B \right)}{\sin 2A-\sin 2B}
Next we will use the formula cos⁡A−cos⁡B=2sin⁡(A+B2)sin⁡(B−A2)\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right) and sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right) above,
=−(2sin⁡(2A+2B2)sin⁡(2B−2A2))2cos⁡(2A+2B2)sin⁡(2A−2B2)= \dfrac{-\left( 2\sin \left( \dfrac{2A+2B}{2} \right)\sin \left( \dfrac{2B-2A}{2} \right) \right)}{2\cos \left( \dfrac{2A+2B}{2} \right)\sin \left( \dfrac{2A-2B}{2} \right)}
=−(2sin⁡(A+B)sin⁡(B−A))2cos⁡(A+B)sin⁡(A−B)= \dfrac{-\left( 2\sin \left( A+B \right)\sin \left( B-A \right) \right)}{2\cos \left( A+B \right)\sin \left( A-B \right)}
Now as we can write sin⁡(B−A)=sin⁡(−(A−B))\sin \left( B-A \right)=\sin \left( -\left( A-B \right) \right) and sin⁡(−θ)=−sin⁡θ\sin \left( -\theta \right)=-\sin \theta using it above we get,
=−(−2sin⁡(A+B)sin⁡(A−B))2cos⁡(A+B)sin⁡(A−B)= \dfrac{-\left( -2\sin \left( A+B \right)\sin \left( A-B \right) \right)}{2\cos \left( A+B \right)\sin \left( A-B \right)}
=2sin⁡(A+B)sin⁡(A−B)2cos⁡(A+B)sin⁡(A−B)= \dfrac{2\sin \left( A+B \right)\sin \left( A-B \right)}{2\cos \left( A+B \right)\sin \left( A-B \right)}
Cancel out the common terms,
=sin⁡(A+B)cos⁡(A+B)= \dfrac{\sin \left( A+B \right)}{\cos \left( A+B \right)}
We know that tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } where θ=A+B\theta =A+B using it above we get,
=tan⁡(A+B)= \tan \left( A+B \right)
This is equal to our RHSRHS term
Hence proved that sin⁡2A−sin⁡2B(sin⁡Acos⁡A)−(sin⁡Bcos⁡B)=tan⁡(A+B)\dfrac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\left( \sin A\cos A \right)-\left( \sin B\cos B \right)}=\tan \left( A+B \right) .

Note:
In this type of question our main aim was to make the numerator in terms of sine function and denominator in terms of cosine function so that we can get our answer in tangent function. As the terms on the left hand side are more, it is more convenient that we solve the left hand side as we can use various identities and relations to eliminate the terms we don’t need.