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Question: If \(\sin A=\dfrac{1}{2},A\in \left( \dfrac{\pi }{2},\pi \right)\) and \(\cos B=\dfrac{\sqrt{3}}{2},...

If sin⁡A=12,A∈(π2,π)\sin A=\dfrac{1}{2},A\in \left( \dfrac{\pi }{2},\pi \right) and cos⁡B=32,B∈(0,π2)\cos B=\dfrac{\sqrt{3}}{2},B\in \left( 0,\dfrac{\pi }{2} \right), then find the value of tan⁡(A+B)\tan \left( A+B \right)

Explanation

Solution

Hint: Use the fact that in the first quadrant tan⁡(x)\tan \left( x \right) is positive and in the second quadrant tan⁡(x)\tan \left( x \right) is negative. Use tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}. Use the fact that if sin⁡x=a\sin x=a, then tan⁡x=±a1−a2\tan x=\pm \dfrac{a}{\sqrt{1-{{a}^{2}}}} and if cos⁡x=a\cos x=a, then tan⁡x=±1−a2a\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⁡2x+sin⁡2x=1{{\cos }^{2}}x+{{\sin }^{2}}x=1 and hence find the value of cos(A+B) using cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B and that of sin⁡(A+B)\sin \left( A+B \right) using sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B

Complete step-by-step answer:
We have sin⁡A=12\sin A=\dfrac{1}{2}
We know that if sin⁡x=a\sin x=a, then tan⁡x=±a1−a2\tan x=\pm \dfrac{a}{\sqrt{1-{{a}^{2}}}}
Hence we have tan⁡A=±121−(12)2=±1232=13\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∈(π2,π)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=−13\tan A=-\dfrac{1}{\sqrt{3}}
Also, we have sin⁡B=32\sin B=\dfrac{\sqrt{3}}{2}
We know that if cos⁡x=a\cos x=a, then tan⁡x=±1−a2a\tan x=\pm \dfrac{\sqrt{1-{{a}^{2}}}}{a}.
Hence we have tan⁡B=±1−(32)232=±1232=±13\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∈(0,π2)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=13\tan B=\dfrac{1}{\sqrt{3}}
Now we know that tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}
Hence we have
tan⁡(A+B)=−13+131−−1313=0\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⁡(A+B)=0\tan \left( A+B \right)=0.

Note: Alternative solution:
We have sin⁡A=12\sin A=\dfrac{1}{2}
Hence sin⁡2A+cos⁡2A=14+cos⁡2A=1{{\sin }^{2}}A+{{\cos }^{2}}A=\dfrac{1}{4}+{{\cos }^{2}}A=1
Hence we have cos⁡A=±32\cos A=\pm \dfrac{\sqrt{3}}{2}
Now since cos is negative in second quadrant, we have cos⁡A=−32\cos A=\dfrac{-\sqrt{3}}{2}
Similarly, we have sin⁡B=12\sin B=\dfrac{1}{2}
Now we have sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B=12×32+(−32)12=0\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⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B=−3232−1212=−1\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⁡(A+B)=sin⁡(A+B)cos⁡(A+B)=0−1=0\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.