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Question: If \(\alpha \) and \(\beta \) be the solution of \(a\cos \theta + b\sin \theta = c\), then 1.\(\si...

If α\alpha and β\beta be the solution of acos⁡θ+bsin⁡θ=ca\cos \theta + b\sin \theta = c, then
1.sin⁡α+sin⁡β=2bca2+b2\sin \alpha + \sin \beta = \dfrac{{2bc}}{{{a^2} + {b^2}}}
2.sin⁡α.sin⁡β=c2−a2a2+b2\sin \alpha .\sin \beta = \dfrac{{{c^2} - {a^2}}}{{{a^2} + {b^2}}}
3.sin⁡α+sin⁡β=2acc2+b2\sin \alpha + \sin \beta = \dfrac{{2ac}}{{{c^2} + {b^2}}}
4.sin⁡α.sin⁡β=a2−b2b2+c2\sin \alpha .\sin \beta = \dfrac{{{a^2} - {b^2}}}{{{b^2} + {c^2}}}

Explanation

Solution

Hint- As we know that acos⁡θ+bsin⁡θ=ca\cos \theta + b\sin \theta = c, then taking bsin⁡θb\sin \theta from the left side of the equation to the right side of the equation we get,
acos⁡θ=c−bsin⁡θa\cos \theta = c - b\sin \theta

Complete step-by-step answer:
Now, squaring both the sides of the equation,

a2cos⁡2θ=(c−bsin⁡θ)2{a^2}{\cos ^2}\theta = {\left( {c - b\sin \theta } \right)^2}

Making the above equation as equation 1,

a2cos⁡2θ=(c−bsin⁡θ)2{a^2}{\cos ^2}\theta = {\left( {c - b\sin \theta } \right)^2} →\to equation 1

Putting the formula of (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab in right side of the equation 1 we get to this result,

a2cos⁡2θ=c2+b2sin⁡2θ−2bcsin⁡θ{a^2}{\cos ^2}\theta = {c^2} + {b^2}{\sin ^2}\theta - 2bc\sin \theta

As we know that cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta , we put this formula in the left side of the above equation to get,

a2(1−sin⁡2θ)=c2+b2sin⁡2θ−2bcsin⁡θ{a^2}\left( {1 - {{\sin }^2}\theta } \right) = {c^2} + {b^2}{\sin ^2}\theta - 2bc\sin \theta

Now we will take the left side of the equation to the right side which will make the left side zero and all the variables will be on the right side. While taking the left side of the equation to the right side we will change the sign i.e. from positive to negative and then solve further,

0=c2+b2sin⁡2θ−2bcsin⁡θ−[a2(1−sin⁡2θ)]0 = {c^2} + {b^2}{\sin ^2}\theta - 2bc\sin \theta - \left[ {{a^2}\left( {1 - {{\sin }^2}\theta } \right)} \right]

The above equation can also be written as,

c2+b2sin⁡2θ−2bcsin⁡θ−[a2(1−sin⁡2θ)]=0{c^2} + {b^2}{\sin ^2}\theta - 2bc\sin \theta - \left[ {{a^2}\left( {1 - {{\sin }^2}\theta } \right)} \right] = 0

Changing the signs, we get the following equation-

c2+b2sin⁡2θ−2bcsin⁡θ−a2+a2sin⁡2θ=0{c^2} + {b^2}{\sin ^2}\theta - 2bc\sin \theta - {a^2} + {a^2}{\sin ^2}\theta = 0

As we can see that sin⁡2θ{\sin ^2}\theta is common in two terms. So, we will take the common out to solve this equation further which will simplify this solution

sin⁡2θ(a2+b2)−2bcsin⁡θ+c2−a2=0{\sin ^2}\theta \left( {{a^2} + {b^2}} \right) - 2bc\sin \theta + {c^2} - {a^2} = 0

Since α\alpha and β\beta are the solutions of the equation the sum of root will be the following equation-

sin⁡α+sin⁡β=2bca2+b2\sin \alpha + \sin \beta = \dfrac{{2bc}}{{{a^2} + {b^2}}}

Which gives us the product of root,

sin⁡α.sin⁡β=c2−a2a2+b2\sin \alpha .\sin \beta = \dfrac{{{c^2} - {a^2}}}{{{a^2} + {b^2}}}

Hence, option 1 and option 2 are the correct options.

Note: Whenever using such questions, always think first about squaring both the sides of the equation so that it will make it easier to put the simple formulae into the equation making the solution easy and fast.