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Question: Which of the following is not correct? (a) \(\dfrac{\left( \sin A+\sin B \right)}{\left( \sin A-\s...

Which of the following is not correct?
(a) (sin⁡A+sin⁡B)(sin⁡A−sin⁡B) = tan⁡(A+B2)tan⁡(A−B2)\dfrac{\left( \sin A+\sin B \right)}{\left( \sin A-\sin B \right)}\ =\ \dfrac{\tan \left( \dfrac{A+B}{2} \right)}{\tan \left( \dfrac{A-B}{2} \right)}
(b) sin⁡2A−cos⁡2B = sin⁡(A+B)sin⁡(A−B){{\sin }^{2}}A-{{\cos }^{2}}B\ =\ \sin \left( A+B \right)\sin \left( A-B \right)
(c) cos⁡A−cos⁡B = 2cos⁡(A+B2)cos⁡(B−A2)\cos A-\cos B\ =\ 2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{B-A}{2} \right)
(d) cos⁡2A−cos⁡2B = sin⁡(A+B)sin⁡(B−A){{\cos }^{2}}A-{{\cos }^{2}}B\ =\ \sin \left( A+B \right)\sin \left( B-A \right)

Explanation

Solution

There are 4 statements given in the question. Take each statement one by one and try to prove it or disprove it. By this you can find out whether the statement is true or false. Use all the required trigonometric identities and also basic knowledge of transformations to get the required result.

Complete step-by-step answer :
Statement-I: Given equation in the question, which we need to solve:
(sin⁡A+sin⁡B)(sin⁡A−sin⁡B) = tan⁡(A+B2)tan⁡(A−B2)\dfrac{\left( \sin A+\sin B \right)}{\left( \sin A-\sin B \right)}\ =\ \dfrac{\tan \left( \dfrac{A+B}{2} \right)}{\tan \left( \dfrac{A-B}{2} \right)}
By taking the left-hand side separately we get the term:
(sin⁡A+sin⁡B)(sin⁡A−sin⁡B)\dfrac{\left( \sin A+\sin B \right)}{\left( \sin A-\sin B \right)}
By basic knowledge of transformation, we can say identities as follows:
sin⁡A+sin⁡B = 2sin⁡(A+B2)cos⁡(A−B2)\sin A+\sin B\ =\ 2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right).
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)
By substituting above terms in the equation’s L.H.S. we get,
L.H.S. = 2sin⁡(A+B2)cos⁡(A−B2)2cos⁡(A+B2)sin⁡(A−B2)=\ \dfrac{2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)}{2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)}
By cancelling 2 and rewriting the term the L.H.S. turns into:
L.H.S. = sin⁡(A+B2)cos⁡(A+B2)sin⁡(A−B2)cos⁡(A−B2)=\ \dfrac{\dfrac{\sin \left( \dfrac{A+B}{2} \right)}{\cos \left( \dfrac{A+B}{2} \right)}}{\dfrac{\sin \left( \dfrac{A-B}{2} \right)}{\cos \left( \dfrac{A-B}{2} \right)}}
By simplifying the above term, we can say
L.H.S. = tan⁡(A+B2)tan⁡(A−B2)=\ \dfrac{\tan \left( \dfrac{A+B}{2} \right)}{\tan \left( \dfrac{A-B}{2} \right)}
L.H.S. == R.H.S.
Hence, the statement is true.
Statement- II: Given statement in the question which we need:
sin⁡2A−cos⁡2B = sin⁡(A+B)sin⁡(A−B){{\sin }^{2}}A-{{\cos }^{2}}B\ =\ \sin \left( A+B \right)\sin \left( A-B \right)
⇒Take R.H.S. sin⁡(A+B)sin⁡(A−B)\Rightarrow \text{Take R.H.S. }\sin \left( A+B \right)\sin \left( A-B \right)
By basic trigonometry knowledge, we know the formulae:
Sin(A + B) = sin A cos B + cos A sin B
Sin(A - B) = sin A cos B – cos A sin B
By substituting above formulae, we get equation as:
= (sin⁡Acos⁡B+cos⁡Asin⁡B)(sin⁡Acos⁡B−cos⁡Asin⁡B)\text{= }\left( \sin A\cos B+\cos A\sin B \right)\left( \sin A\cos B-\cos A\sin B \right)
By using algebraic identity, given by:
(a+b)(a-b) = a2−b2{{a}^{2}}-{{b}^{2}}
By substituting above identity, we get the equation as:
=sin⁡2Acos⁡2B−cos⁡2Asin⁡2B={{\sin }^{2}}A{{\cos }^{2}}B-{{\cos }^{2}}A{{\sin }^{2}}B
By using the identity: sin⁡2A+cos⁡2A = 1{{\sin }^{2}}A+{{\cos }^{2}}A\ =\ 1, we can write sin⁡2A= 1−cos⁡2A{{\sin }^{2}}A=\ 1-{{\cos }^{2}}A
By substituting the above, we get it as:
=sin⁡2A(1−sin⁡2B)−sin⁡2B(1−sin⁡2A)={{\sin }^{2}}A\left( 1-{{\sin }^{2}}B \right)-{{\sin }^{2}}B\left( 1-{{\sin }^{2}}A \right)
By simplifying the above equation, we get it as:
=sin⁡2A−sin⁡2Asin⁡2B−sin⁡2B+sin⁡2Asin⁡2B = sin⁡2A−sin⁡2B={{\sin }^{2}}A-{{\sin }^{2}}A{{\sin }^{2}}B-{{\sin }^{2}}B+{{\sin }^{2}}A{{\sin }^{2}}B\ =\ {{\sin }^{2}}A-{{\sin }^{2}}B
L.H.S. = sin⁡2A−cos⁡2B {{\sin }^{2}}A-{{\cos }^{2}}B\
R.H.S. = sin⁡2A−sin⁡2B{{\sin }^{2}}A-{{\sin }^{2}}B
L.H.S. ≠\ne R.H.S. so this is wrong statement.
Statement-III: Given statement in question which we need is
cos⁡A−cos⁡B = 2cos⁡(A+B2)cos⁡(A−B2)\cos A-\cos B\ =\ 2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)
L.H.S. = cos A – cos B
By transformations, we know
cos⁡A−cos⁡B = −2sin⁡(A+B2)sin⁡(A−B2)\cos A-\cos B\ =\ -2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)
So from above equation we get:
L.H.S. =−2sin⁡(A+B2)sin⁡(A−B2)-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)
From the given equation, we get:
R.H.S. = 2cos⁡(A+B2)cos⁡(A−B2)2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)
L.H.S. ≠\ne R.H.S.
So, This statement is wrong.
Statement-IV: Given statement in question, which we need is
cos⁡2A−cos⁡2B = sin⁡(A+B)sin⁡(B−A){{\cos }^{2}}A-{{\cos }^{2}}B\ =\ \sin \left( A+B \right)\sin \left( B-A \right)
By using algebraic identity, given by:
(a+b)(a-b) = a2−b2{{a}^{2}}-{{b}^{2}}
By substituting above identity, we get the left hand side as:
⇒(cos⁡A−cos⁡B)(cos⁡A+cos⁡B)\Rightarrow \left( \cos A-\cos B \right)\left( \cos A+\cos B \right)
By transformations, we can say that:
(cos⁡A−cos⁡B)\left( \cos A-\cos B \right)= −2sin⁡(A−B2)sin⁡(A+B2)-2\sin \left( \dfrac{A-B}{2} \right)\sin \left( \dfrac{A+B}{2} \right)
(cos⁡A+cos⁡B)\left( \cos A+\cos B \right)= 2cos⁡(A+B2)cos⁡(A+B2)2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)
By substituting these values into equation above, we get the left hand side as:
⇒−2sin⁡(A−B2)sin⁡(A+B2)⋅2cos⁡(A+B2)cos⁡(A+B2)\Rightarrow -2\sin \left( \dfrac{A-B}{2} \right)\sin \left( \dfrac{A+B}{2} \right)\centerdot 2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)
⇒−2sin⁡(A−B2)cos⁡(A−B2)⋅2sin⁡(A+B2)cos⁡(A+B2)\Rightarrow -2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\centerdot 2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)
By using 2sin(A)cos⁡(A) = sin⁡2A\text{2sin}\left( A \right)\cos \left( A \right)\ =\ \sin 2A
We get the left hand side ⇒−sin⁡(2.A−B2)sin⁡(2.A+B2) = sin⁡(B−A)sin⁡(A+B)\Rightarrow -\sin \left( 2.\dfrac{A-B}{2} \right)\sin \left( 2.\dfrac{A+B}{2} \right)\ =\ \sin \left( B-A \right)\sin \left( A+B \right)
So from this, we get:
L.H.S. = sin⁡(B−A)sin⁡(A+B)\sin \left( B-A \right)\sin \left( A+B \right)
From given statement, we can say:
R.H.S. = sin⁡(B−A)sin⁡(A+B)\sin \left( B-A \right)\sin \left( A+B \right)
So, L.H.S. = R.H.S.
Statement is correct. So, statements II, III are wrong.
Option (b), (c) is correct.

Note :While solving statement 1 be careful that as the terms in R.H.S. are in terms of tan you must convert L.H.S. into terms of sin⁡cos⁡\dfrac{\sin }{\cos }but not cos⁡sin⁡\dfrac{\cos }{\sin } because the latter will give everything in terms of cot but we need tan. Take care of the “−-” sign in the transformation of cos⁡a−cos⁡b\cos a-\cos b as this may make the statement IV wrong. Don’t forget that “−-” sign in the transformation.