Question & Answer
QUESTION

If \[\sin A+\cos B=a\text{ and }\sin B+\cos A=b\], then \[\sin \left( A+B \right)\] is equal to:
(a) \[\dfrac{{{a}^{2}}+{{b}^{2}}}{2}\]
(b) \[\dfrac{{{a}^{2}}-{{b}^{2}}+2}{2}\]
(c) \[\dfrac{{{a}^{2}}+{{b}^{2}}-2}{2}\]
(d) None of these

ANSWER Verified Verified
Hint: In this question, we first need to square the given two equations and then add them both. Then by using the trigonometric identity and by using the formula trigonometric ratios for compound angles we can get the result.

Complete step by step answer:
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
\[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\]
From the equations given in the question we have,
\[\sin A+\cos B=a........\left( 1 \right)\]
\[\sin B+\cos A=b.........\left( 2 \right)\]
Now, on considering the equation (1) above and squaring it on both sides we get,
\[\Rightarrow {{\left( \sin A+\cos B \right)}^{2}}={{a}^{2}}\]
Let us now expand the terms on the left hand side then we get,
\[\Rightarrow {{\sin }^{2}}A+{{\cos }^{2}}B+2\sin A\cos B={{a}^{2}}.....\left( 3 \right)\]
Now, on considering the equation (2) and squaring it on both sides we get,
\[\Rightarrow {{\left( \sin B+\cos A \right)}^{2}}={{b}^{2}}\]
Let us now expand the terms on the left hand side then we get,
\[\Rightarrow {{\sin }^{2}}B+{{\cos }^{2}}A+2\sin B\cos A={{b}^{2}}.....\left( 4 \right)\]
As we already know that from the trigonometric identities that,
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
From, trigonometric ratios of compound angles we know that the algebraic sum of two or more angles are called compound angles which gives us the formula:
\[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\]
Let us now consider the above equation (3) and equation (4)
Now, on adding the equation (3) and equation (4) we get,
\[\Rightarrow {{\sin }^{2}}A+{{\cos }^{2}}B+2\sin A\cos B+{{\sin }^{2}}B+{{\cos }^{2}}A+2\sin B\cos A={{a}^{2}}+{{b}^{2}}\]
Now, on writing the sine and cosine terms of same angle together we get,
\[\Rightarrow {{\sin }^{2}}A+{{\cos }^{2}}A+{{\sin }^{2}}B+{{\cos }^{2}}B+2\sin A\cos B+2\sin B\cos A={{a}^{2}}+{{b}^{2}}\]
Now, by using the trigonometric identity and on rewriting the terms in the above equation we get,
\[\Rightarrow 1+1+2\sin A\cos B+2\sin B\cos A={{a}^{2}}+{{b}^{2}}\]
Now, on further simplification we get,
\[\Rightarrow 2+2\sin A\cos B+2\sin B\cos A={{a}^{2}}+{{b}^{2}}\]
Now, by subtracting 2 on both sides we get,
\[\Rightarrow 2\sin A\cos B+2\sin B\cos A={{a}^{2}}+{{b}^{2}}-2\]
Let us now divide the above equation with 2 on both sides,
\[\Rightarrow \sin A\cos B+\sin B\cos A=\dfrac{{{a}^{2}}+{{b}^{2}}-2}{2}\]
Now, by using the formula for trigonometric ratios of compound angles the above equation can be further written as.
\[\therefore \sin \left( A+B \right)=\dfrac{{{a}^{2}}+{{b}^{2}}-2}{2}\]
Hence, the correct option is (c).

Note: Instead of squaring the given two equations and then adding them to get the required form we can also first expand the \[\sin \left( A+B \right)\] and then substitute the required values from the given equations. But this method would be a little confusing.
It is important to note that while applying the trigonometric identity the angles of both cosine and sine terms should be the same then only it will be 1. After squaring one equation we cannot write the sum of the squares of the sine and cosine terms as 1 because the angles are different.
\[{{\sin }^{2}}B+{{\cos }^{2}}A\ne 1\]