Question

If $\sin A=\dfrac{1}{2},A\in \left( \dfrac{\pi }{2},\pi \right)$ and $\cos B=\dfrac{\sqrt{3}}{2},B\in \left( 0,\dfrac{\pi }{2} \right)$, then find the value of $\tan \left( A+B \right)$

Answer 1

Hint: Use the fact that in the first quadrant $\tan \left( x \right)$ is positive and in the second quadrant $\tan \left( x \right)$ is negative. Use $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$. Use the fact that if $\sin x=a$, then $\tan x=\pm \dfrac{a}{\sqrt{1-{{a}^{2}}}}$ and if $\cos x=a$, then $\tan x=\pm \dfrac{\sqrt{1-{{a}^{2}}}}{a}$. Hence find the value of tanA and tanB and hence find the value of tan(A+B). Alternatively, find the values of cos(A) and cos(B) using ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ and hence find the value of cos(A+B) using $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ and that of $\sin \left( A+B \right)$ using $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$

Complete step-by-step answer:
We have $\sin A=\dfrac{1}{2}$
We know that if $\sin x=a$, then $\tan x=\pm \dfrac{a}{\sqrt{1-{{a}^{2}}}}$
Hence we have $\tan A=\pm \dfrac{\dfrac{1}{2}}{\sqrt{1-{{\left( \dfrac{1}{2} \right)}^{2}}}}=\pm \dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}=\dfrac{1}{\sqrt{3}}$
But since $A\in \left( \dfrac{\pi }{2},\pi \right)$, we have A lies in the second quadrant and since in second quadrant tan is negative, we have
$\tan A=-\dfrac{1}{\sqrt{3}}$
Also, we have $\sin B=\dfrac{\sqrt{3}}{2}$
We know that if $\cos x=a$, then $\tan x=\pm \dfrac{\sqrt{1-{{a}^{2}}}}{a}$.
Hence we have $\tan B=\pm \dfrac{\sqrt{1-{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}}{\dfrac{\sqrt{3}}{2}}=\pm \dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}=\pm \dfrac{1}{\sqrt{3}}$
But since $B\in \left( 0,\dfrac{\pi }{2} \right)$, B lies in the first quadrant.
Since in first quadrant tan is positive, we have
$\tan B=\dfrac{1}{\sqrt{3}}$
Now we know that $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
Hence we have
$\tan \left( A+B \right)=\dfrac{-\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}}{1-\dfrac{-1}{\sqrt{3}}\dfrac{1}{\sqrt{3}}}=0$
Hence $\tan \left( A+B \right)=0$.

Note: Alternative solution:
We have $\sin A=\dfrac{1}{2}$
Hence ${{\sin }^{2}}A+{{\cos }^{2}}A=\dfrac{1}{4}+{{\cos }^{2}}A=1$
Hence we have $\cos A=\pm \dfrac{\sqrt{3}}{2}$
Now since cos is negative in second quadrant, we have \[\cos A=\dfrac{-\sqrt{3}}{2}\]
Similarly, we have $\sin B=\dfrac{1}{2}$
Now we have $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B=\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}+\left( \dfrac{-\sqrt{3}}{2} \right)\dfrac{1}{2}=0$
Also, $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B=\dfrac{-\sqrt{3}}{2}\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}\dfrac{1}{2}=-1$
Hence $\tan \left( A+B \right)=\dfrac{\sin \left( A+B \right)}{\cos \left( A+B \right)}=\dfrac{0}{-1}=0$, which is the same as obtained above,
[2] You can remember the sign of each trigonometric ratio in each quadrant with the mnemonic:
“Add Sugar To Coffee”
Add: All positive in the first quadrant
Sugar: Sine and Cosine(Inverse of sine) positive in the second quadrant.
To: Tangent and Cotangent positive in the third quadrant.
Coffee: cosine and secant positive in the fourth quadrant.