Question

If $\sin A=x $ , $\cos B=y $ and $A+B+C=0 $, then ${{x}^{2}}+2xy\sin C+{{y}^{2}} $is equal to
(A) ${{\sin }^{2}}C $
(B) ${{\cos }^{2}}C $
(C) $1+{{\sin }^{2}}C $
(D) $1+{{\cos }^{2}}C $

Answer 1

Hint: For answering this question we will use the given values $\sin A=x $, $\cos B=y $ and $A+B+C=0 $in the given equation ${{x}^{2}}+2xy\sin C+{{y}^{2}} $ and we will simplify it using $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B $ and $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B $.

Complete step by step answer:
From the question we have $\sin A=x $, $\cos B=y $and $A+B+C=0 $.
From $A+B=C=0 $we have $C=-\left( A+B \right) $, $C=-\left( A+B \right) $, by using this we will have $\sin C=\sin \left[ -\left( A+B \right) \right] $ .
Let us substitute the respective values in the given equation ${{x}^{2}}+2xy\sin C+{{y}^{2}} $. After that we will have ${{\sin }^{2}}A+2\sin A\cos B\sin \left[ -\left( A+B \right) \right]+{{\cos }^{2}}B $ .
Since $\sin \left( -\theta \right)=-\sin \theta $ by using this we will have ${{\sin }^{2}}A-2\sin A\cos B\sin \left( A+B \right)+{{\cos }^{2}}B $ .
Let us expand the term using the expansion $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B $ . After expanding we will have ${{\sin }^{2}}A-2\sin A\cos B\left[ \sin A\cos B+\cos A\sin B \right]+{{\cos }^{2}}B $ .
Let us expand this ${{\sin }^{2}}A-2\sin A\cos B\sin A\cos B-2\sin A\cos B\cos A\sin A+{{\cos }^{2}}B $
If we observe we have $\sin A\cos B $ appeared 2 times in a single term this can be written as square. This can be mathematically given as ${{\sin }^{2}}A-2{{\sin }^{2}}A{{\cos }^{2}}B-2\sin A\cos B\cos A\sin A+{{\cos }^{2}}B $ .
Let us expand the term $2{{\sin }^{2}}A{{\cos }^{2}}B $ after this we will have ${{\sin }^{2}}A-{{\sin }^{2}}A{{\cos }^{2}}B-{{\sin }^{2}}A{{\cos }^{2}}B-2\sin A\cos B\cos A\sin A+{{\cos }^{2}}B $ .
If we observe we have ${{\sin }^{2}}A $ and ${{\cos }^{2}}A $ are common in 2 terms if we take them out we will have ${{\sin }^{2}}A\left( 1-{{\cos }^{2}}B \right)-2\sin A\cos B\cos A\sin A+{{\cos }^{2}}B\left( 1-{{\sin }^{2}}A \right) $ .
Since we have ${{\sin }^{2}}x+{{\cos }^{2}}x=1 $ . By using and simplifying this we will have ${{\sin }^{2}}A{{\sin }^{2}}B-2\sin A\cos B\cos A\sin A+{{\cos }^{2}}B{{\cos }^{2}}A $ .
If we observe we have $\sin A\cos B\cos A\sin B $ appeared 2 times in a single term this can be written as square. This can be mathematically given as ${{\sin }^{2}}A{{\sin }^{2}}B-\sin A\cos B\cos A\sin B-\sin A\cos B\cos A\sin B+{{\cos }^{2}}B{{\cos }^{2}}A $ .
If we observe we have $\sin A\sin B $ and $\cos A\cos B $ are common in 2 terms if we take them out we will have $\sin A\sin B\left( \sin A\sin B-\cos A\cos B \right)+\cos B\cos A\left( \cos A\cos B-\sin A\sin B \right) $.
If we observe we have $\left( \sin A\sin B-\cos A\cos B \right) $ are common in 2 terms if we take them out we will have $\left( \sin A\sin B-\cos A\cos B \right)\left( \sin A\sin B-\cos A\cos B \right) $ .
Since we know that $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B $ by using this expansion we will have $\left( -\cos \left( A+B \right) \right)\left( -\cos \left( A+B \right) \right) $ .
Since we know that $\text{negative}\times negative=positive $ we will have $\left( \cos \left( A+B \right) \right)\left( \cos \left( A+B \right) \right) $ .
If we observe we have $\cos \left( A+B \right) $ appeared 2 times in a single term this can be written as square. This can be mathematically given as ${{\cos }^{2}}\left( A+B \right) $ .
From $A+B=C=0 $we have $C=-\left( A+B \right) $, by using this we will have ${{\cos }^{2}}\left( A+B \right)={{\cos }^{2}}\left( -C \right) $ .
Since $\cos \left( -\theta \right)=+\cos \theta $ by using this we will have ${{\cos }^{2}}C $ .
So we will end up with a conclusion that when $\sin A=x $ , $\cos B=y $ and $A+B+C=0 $, then ${{x}^{2}}+2xy\sin C+{{y}^{2}}={{\cos }^{2}}C $ .

So, the correct answer is “Option B”.

Note: While solving questions of this type we should remember that the $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B $ not $\cos \left( A+B \right)=\cos A\cos B+\sin A\sin B $ if we use this by mistake we will end up having the conclusion ${{x}^{2}}+2xy\sin C+{{y}^{2}}={{\cos }^{2}}\left( A-B \right) $ . It is completely a wrong answer.