Question

If in a \[\Delta ABC\], \[\angle C={{90}^{o}}\], then prove that \[\sin \left( A-B \right)=\dfrac{\left( {{a}^{2}}-{{b}^{2}} \right)}{\left( {{a}^{2}}+{{b}^{2}} \right)}\].

Answer 1

Hint: In the question, it is given that the \[\Delta ABC\] is a right-angled triangle at C. We know that in a right-angled triangle, the sum of the square of the hypotenuse is equal to the sum of the squares of the other two sides and this is known as Pythagoras theorem. Also, we will use the difference angle identity of sine, sine rule of the triangle, and cosine rule of the triangle. The identity of the difference angle of sine is as follows:
\[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\]

Complete step by step answer:
We have been given that \[\angle C={{90}^{o}}\]. So, by using Pythagoras theorem, we get,
\[{{a}^{2}}+{{b}^{2}}={{c}^{2}}.....\left( i \right)\]
We have to prove that \[\sin \left( A-B \right)=\dfrac{\left( {{a}^{2}}-{{b}^{2}} \right)}{\left( {{a}^{2}}+{{b}^{2}} \right)}\]
Let us take the left-hand side of the equality and by using the identity, we get,
\[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B.....\left( ii \right)\]
We know that the sine rule of the triangle is
\[\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\]
Let us assume the ratio as ‘k’.
\[\Rightarrow \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=k\]
Also, we have \[\angle C={{90}^{o}}\],
\[\Rightarrow \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin 90}{c}=k\]
\[\Rightarrow \sin A=ak,\text{ }\sin B=bk\text{ and }\sin 90=ck=1\]
Also, we get, \[k=\dfrac{1}{c}\]
By substituting these values in the equation (ii), we get,
\[\sin \left( A-B \right)=ak\cos B-\cos A.bk\]
We know that the cosine rule of the triangle is as follows:
\[\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\]
\[\cos B=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\]
By substituting these values in the equation, we get,
\[\sin \left( A-B \right)=ak.\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}-\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}.bk\]
On simplification, we get,
\[\sin \left( A-B \right)=\dfrac{k\left( {{a}^{2}}+{{c}^{2}}-{{b}^{2}} \right)}{2c}-\dfrac{k\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)}{2c}\]
Taking \[\dfrac{k}{2c}\] as common, we get,
\[\sin \left( A-B \right)=\dfrac{k}{2c}\left[ {{a}^{2}}+{{c}^{2}}-{{b}^{2}}-{{b}^{2}}-{{c}^{2}}+{{a}^{2}} \right]\]
\[\Rightarrow \sin \left( A-B \right)=\dfrac{k}{2c}\left( 2{{a}^{2}}-2{{b}^{2}} \right)\]
\[\Rightarrow \sin \left( A-B \right)=\dfrac{k}{2c}\times 2\left( {{a}^{2}}-{{b}^{2}} \right)\]
\[\Rightarrow \sin \left( A-B \right)=\dfrac{k}{c}\left( {{a}^{2}}-{{b}^{2}} \right)\]
Using the value \[k=\dfrac{1}{c}\], we get,
\[\Rightarrow \sin \left( A-B \right)=\dfrac{1}{{{c}^{2}}}\left( {{a}^{2}}-{{b}^{2}} \right)\]
Again, using the equation (i), we get,
\[\sin \left( A-B \right)=\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\]
Hence, it is proved that \[\sin \left( A-B \right)=\dfrac{\left( {{a}^{2}}-{{b}^{2}} \right)}{\left( {{a}^{2}}+{{b}^{2}} \right)}\].

Note: In these types of questions, we should approach by using sine rule and cosine rule. We can also prove the given expression by taking the right-hand side of the equality and using the sine and cosine rule of the triangle.