Question
$$A + iB$$ form of $$\frac{{\left( {\cos x + i\sin x} \right)\left( {\cos y + i\sin y} \right)}}{{\left( {\cot u + i} \right)\left( {1 + i\tan v} \right)}}$$ is equal to :
A.
$$\sin u\cos v\left[ {\cos \left( {x + y - u - v} \right) + i\sin \left( {x + y - u - v} \right)} \right]$$
B.
$$\sin u\cos v\left[ {\cos \left( {x + y + u + v} \right) + i\sin \left( {x + y + u + v} \right)} \right]$$
C.
$$\sin u\cos v\left[ {\cos \left( {x + y + u + v} \right) - i\sin \left( {x + y - u + v} \right)} \right]$$
D.
None of these
Answer :
$$\sin u\cos v\left[ {\cos \left( {x + y - u - v} \right) + i\sin \left( {x + y - u - v} \right)} \right]$$
Solution :
$$\eqalign{
& {\text{Given, }}\frac{{\left( {\cos x + i\sin x} \right)\left( {\cos y + i\sin y} \right)}}{{\left( {\cot u + i} \right)\left( {1 + i\tan v} \right)}} \cr
& = \frac{{\left( {\cos x + i\sin x} \right)\left( {\cos y + i\sin y} \right)}}{{\left( {\cos u + i\sin u} \right)\left( {\cos v + i\sin v} \right)}} \cr
& = \sin u\cos v\left[ {\cos \left( {x + y - u - v} \right) + i\sin \left( {x + y - u - v} \right)} \right] \cr} $$