MCQ
If $\cos(\text{A}-\text{B})=\frac35$ and $\tan\text{A}\tan\text{B}=2,$ then
- A$\cos\text{A}\cos\text{B}=\frac15$
- B$\cos\text{A}\cos\text{B}=-\frac15$
- C$\sin\text{A}\sin\text{B}=-\frac15$
- D$\sin\text{A}\sin\text{B}=-\frac15$
Solution:
$\tan\text{A}\tan\text{B}=\frac{\sin\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}=2\ (\text{Given})\cdots(1)$
Also,
$\cos(\text{A - B})=\frac35$
$\Rightarrow\cos\text{A}\cos\text{B}=\frac35+\sin\text{A}\sin\text{A}=\frac35$
$\therefore\sin\text{A}\sin\text{B}=\frac35-\cos\text{A}\cos\text{B}\cdots(2)$
Substituting eq (2) in eq (1), we get:
$\Rightarrow\frac{\frac35-\cos\text{A}\cos\text{B}}{\cos\text{A}\cos\text{B}}=2$
$\Rightarrow3\cos\text{A}\cos\text{B}=\frac35$
$\Rightarrow\cos\text{A}\cos\text{B}=\frac15$
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