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Question: If \[\sin (\theta )+\sin (\phi )=a\] and \[\cos (\theta )+\cos (\phi )=b\], then \[\tan \left( \dfra...

If sin⁡(θ)+sin⁡(ϕ)=a\sin (\theta )+\sin (\phi )=a and cos⁡(θ)+cos⁡(ϕ)=b\cos (\theta )+\cos (\phi )=b, then tan⁡(θ−ϕ2)\tan \left( \dfrac{\theta -\phi }{2} \right) is equal to
A) a2+b24−a2−b2\sqrt{\dfrac{{{a}^{2}}+{{b}^{2}}}{4-{{a}^{2}}-{{b}^{2}}}}
B) 4−a2−b2a2+b2\sqrt{\dfrac{4-{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}}
C) a2+b24+a2+b2\sqrt{\dfrac{{{a}^{2}}+{{b}^{2}}}{4+{{a}^{2}}+{{b}^{2}}}}
D) 4+a2+b2a2+b2\sqrt{\dfrac{4+{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}}

Explanation

Solution

According to the given question here there are two equation we have to squaring and adding those equation and using trigonometry identity formula that is sin⁡2(θ)+cos⁡2(ϕ)=1{{\sin }^{2}}(\theta )+{{\cos }^{2}}(\phi )=1 and also apply formula of cos⁡(θ−ϕ)=cos⁡(θ)cos⁡(ϕ)+sin⁡(θ)sin⁡(ϕ)\cos (\theta -\phi )=\cos (\theta )\cos (\phi )+\sin (\theta )\sin (\phi ) and solve this problem.

Complete step by step answer:
According to the given condition there are two equation which is given:
That is, sin⁡(θ)+sin⁡(ϕ)=a−−−−(1)\sin (\theta )+\sin (\phi )=a----(1)
cos⁡(θ)+cos⁡(ϕ)=b−−−−(2)\cos (\theta )+\cos (\phi )=b----(2)
We have to find the value of tan⁡(θ−ϕ2)\tan \left( \dfrac{\theta -\phi }{2} \right)
To find this we have two equations individually by squaring the equation (1) and equation (2).
First of all we will solve the first equation
By squaring on both side we get:
(sin⁡(θ)+sin⁡(ϕ))2=a2{{\left( \sin (\theta )+\sin (\phi ) \right)}^{2}}={{a}^{2}}
So, by using this property of (a+b)2=a2+2ab+b2{{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}we get:
sin⁡2(θ)+sin⁡2(ϕ)+2sin⁡(θ)sin⁡(ϕ)=a2−−−(3){{\sin }^{2}}(\theta )+{{\sin }^{2}}(\phi )+2\sin (\theta )\sin (\phi )={{a}^{2}}---(3)
Similarly, we can also perform for equation (2).
By squaring the equation (2)
(cos⁡(θ)+cos⁡(ϕ))2=b2{{\left( \cos (\theta )+\cos (\phi ) \right)}^{2}}={{b}^{2}}
By simplifying this equation by using the property of (a+b)2=a2+ab+b2{{(a+b)}^{2}}={{a}^{2}}+ab+{{b}^{2}}we get:
cos⁡2(θ)+cos⁡2(ϕ)+2cos⁡(θ)cos⁡(ϕ)=b2−−(4){{\cos }^{2}}(\theta )+{{\cos }^{2}}(\phi )+2\cos (\theta )\cos (\phi )={{b}^{2}}--(4)
Add this equation (3) and equation (4)
sin⁡2(θ)+sin⁡2(ϕ)+2sin⁡(θ)sin⁡(ϕ)+cos⁡2(θ)+cos⁡2(ϕ)+2cos⁡(θ)cos⁡(ϕ)=a2+b2{{\sin }^{2}}(\theta )+{{\sin }^{2}}(\phi )+2\sin (\theta )\sin (\phi )+{{\cos }^{2}}(\theta )+{{\cos }^{2}}(\phi )+2\cos (\theta )\cos (\phi )={{a}^{2}}+{{b}^{2}}
By arranging the term we get:
sin⁡2(θ)+cos⁡2(θ)+sin⁡2(ϕ)+cos⁡2(ϕ)+2sin⁡(θ)sin⁡(ϕ)+2cos⁡(θ)cos⁡(ϕ)=a2+b2{{\sin }^{2}}(\theta )+{{\cos }^{2}}(\theta )+{{\sin }^{2}}(\phi )+{{\cos }^{2}}(\phi )+2\sin (\theta )\sin (\phi )+2\cos (\theta )\cos (\phi )={{a}^{2}}+{{b}^{2}}
By using the property sin⁡2(θ)+cos⁡2(θ)=1{{\sin }^{2}}(\theta )+{{\cos }^{2}}(\theta )=1and simplify it
1+1+2sin⁡(θ)sin⁡(ϕ)+2cos⁡(θ)cos⁡(ϕ)=a2+b21+1+2\sin (\theta )\sin (\phi )+2\cos (\theta )\cos (\phi )={{a}^{2}}+{{b}^{2}}
2+2sin⁡(θ)sin⁡(ϕ)+2cos⁡(θ)cos⁡(ϕ)=a2+b22+2\sin (\theta )\sin (\phi )+2\cos (\theta )\cos (\phi )={{a}^{2}}+{{b}^{2}}
Take the 2 common on LHS and use the trigonometric property that iscos⁡(θ−ϕ)=cos⁡(θ)cos⁡(ϕ)+sin⁡(θ)sin⁡(ϕ)\cos (\theta -\phi )=\cos (\theta )\cos (\phi )+\sin (\theta )\sin (\phi )
2(1+cos⁡(θ)cos⁡(ϕ)+sin⁡(θ)sin⁡(ϕ))=a2+b22(1+\cos (\theta )\cos (\phi )+\sin (\theta )\sin (\phi ))={{a}^{2}}+{{b}^{2}}
2(1+cos⁡(θ−ϕ2))=a2+b22\left( 1+\cos \left( \dfrac{\theta -\phi }{2} \right) \right)={{a}^{2}}+{{b}^{2}}
By using the property of half angle formula we get:
4cos⁡2(θ−ϕ2)=a2+b24{{\cos }^{2}}\left( \dfrac{\theta -\phi }{2} \right)={{a}^{2}}+{{b}^{2}}
By simplifying further we get:
cos⁡2(θ−ϕ2)=a2+b24{{\cos }^{2}}\left( \dfrac{\theta -\phi }{2} \right)=\dfrac{{{a}^{2}}+{{b}^{2}}}{4}
cos⁡(θ−ϕ2)=a2+b24−−(5)\cos \left( \dfrac{\theta -\phi }{2} \right)=\sqrt{\dfrac{{{a}^{2}}+{{b}^{2}}}{4}}--(5)
Now, we have to use again trigonometric identity sin⁡2(θ−ϕ2)=1−cos⁡2(θ−ϕ2){{\sin }^{2}}\left( \dfrac{\theta -\phi }{2} \right)=1-{{\cos }^{2}}\left( \dfrac{\theta -\phi }{2} \right)we get:
sin⁡2(θ−ϕ2)=1−a2+b24{{\sin }^{2}}\left( \dfrac{\theta -\phi }{2} \right)=1-\dfrac{{{a}^{2}}+{{b}^{2}}}{4}
After simplifying we get:
sin⁡2(θ−ϕ2)=4−a2−b24{{\sin }^{2}}\left( \dfrac{\theta -\phi }{2} \right)=\dfrac{4-{{a}^{2}}-{{b}^{2}}}{4}
Take root on both sides we get:
sin⁡(θ−ϕ2)=4−a2−b24−−(6)\sin \left( \dfrac{\theta -\phi }{2} \right)=\sqrt{\dfrac{4-{{a}^{2}}-{{b}^{2}}}{4}}--(6)
To find the value of tan⁡(θ−ϕ2)\tan \left( \dfrac{\theta -\phi }{2} \right)
We have to use the formula for
tan⁡(θ−ϕ2)=sin⁡(θ−ϕ2)cos⁡(θ−ϕ2)−−−(7)\tan \left( \dfrac{\theta -\phi }{2} \right)=\dfrac{\sin \left( \dfrac{\theta -\phi }{2} \right)}{\cos \left( \dfrac{\theta -\phi }{2} \right)}---(7)
Substitute the value of equation (5) and equation (6) in equation (7)
tan⁡(θ−ϕ2)=4−a2−b24a2+b24\tan \left( \dfrac{\theta -\phi }{2} \right)=\dfrac{\sqrt{\dfrac{4-{{a}^{2}}-{{b}^{2}}}{4}}}{\sqrt{\dfrac{{{a}^{2}}+{{b}^{2}}}{4}}}
After simplifying this equation and by solving 4 get cancelled we get:
tan⁡(θ−ϕ2)=4−a2−b2a2+b2\tan \left( \dfrac{\theta -\phi }{2} \right)=\dfrac{\sqrt{4-{{a}^{2}}-{{b}^{2}}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}
By taking root common from denominator as well as numerator we get:
tan⁡(θ−ϕ2)=4−a2−b2a2+b2\tan \left( \dfrac{\theta -\phi }{2} \right)=\sqrt{\dfrac{4-{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}}
Therefore, the correct option is option (B).

Note:
According to this particular problem we have to remember that we have to apply the correct identity formula of trigonometry. This approach to the problem is very important and proper trigonometry formulas to be remembered. Don’t confuse yourself in applying the formula and simplifying the problems. So, the above solution is referred for solving the problems.