Question

Let $\sin (A+B)=1$ and $\sin (A-B)=1/2$, where $A,Bϵ[0,\frac{\pi }{2}]$ .

What is $\tan (A+2B).\tan (2A+B)$ equal to ?

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (C)

1

We know that $\sin \left({90}^o\right)=1$ and $\sin \left({30}^o\right)=0.5$

So, from the given two equations we get,

$A+B={90}^o$ .... $\left(i\right)$ and $A-B={30}^o$ ..... $\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$, we get

$A={60}^o$ and $B={30}^o$

So, $\tan (A+2B)=\tan ({60}^o+2\times {30}^o)=\tan \left({120}^o\right)=-\cot \left({30}^o\right)=-\sqrt{3}$

And $\tan (2A+B)=\tan (2\times {60}^o+{30}^o)=\tan \left({150}^o\right)=-\cot \left({60}^0\right)=\frac{-1}{\sqrt{3}}$

So, $\tan (A+2B).\tan (2A+B)=-\sqrt{3}\times \frac{-1}{\sqrt{3}}=1$