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Question: Eliminate ‘ \( \theta \) ’ from the equation and find the eliminant: \( x\cos \theta + y\sin \the...

Eliminate ‘ θ\theta ’ from the equation and find the eliminant:
xcos⁡θ+ysin⁡θ=cx\cos \theta + y\sin \theta = c and xcos⁡θ−ysin⁡θ=dx\cos \theta - y\sin \theta = d

Explanation

Solution

Hint : The two equations can be added up and one term will get cancelled this will eliminate the sine term in the equation. Now we will use the value of the cosine term and proceed further to solve the system of equations. The simple technique is to square the equations and add them. Then we will use the formulas written here and reduce the equation to a simpler form.
Formula used:
A general trigonometric identity is sin⁡2θ+cos⁡2θ=1⇒sin⁡2θ=1−cos⁡2θ{\sin ^2}\theta + {\cos ^2}\theta = 1 \Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta
And 2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta = \sin 2\theta

Complete step-by-step answer :
The given equations are xcos⁡θ+ysin⁡θ=cx\cos \theta + y\sin \theta = c and xcos⁡θ−ysin⁡θ=dx\cos \theta - y\sin \theta = d
Adding the two equations we get,
xcos⁡θ+ysin⁡θ+xcos⁡θ−ysin⁡θ=c+d⇒2xcos⁡θ=c+d⇒cos⁡θ=c+d2xx\cos \theta + y\sin \theta + x\cos \theta - y\sin \theta = c + d \Rightarrow 2x\cos \theta = c + d \Rightarrow \cos \theta = \dfrac{{c + d}}{{2x}} --(1)
On squaring them both we get,
(xcos⁡θ+ysin⁡θ)2=c2{\left( {x\cos \theta + y\sin \theta } \right)^2} = {c^2} and (xcos⁡θ−ysin⁡θ)2=d2{\left( {x\cos \theta - y\sin \theta } \right)^2} = {d^2}
Now on simplifying we have,
x2cos⁡2θ+y2sin⁡2θ+2xycos⁡θsin⁡θ=c2{x^2}{\cos ^2}\theta + {y^2}{\sin ^2}\theta + 2xy\cos \theta \sin \theta = {c^2} and x2cos⁡2θ+y2sin⁡2θ−2xycos⁡θsin⁡θ=d2{x^2}{\cos ^2}\theta + {y^2}{\sin ^2}\theta - 2xy\cos \theta \sin \theta = {d^2}
⇒x2cos⁡2θ+y2sin⁡2θ+xysin⁡2θ=c2\Rightarrow {x^2}{\cos ^2}\theta + {y^2}{\sin ^2}\theta + xy\sin 2\theta = {c^2} and x2cos⁡2θ+y2sin⁡2θ−xysin⁡2θ=d2{x^2}{\cos ^2}\theta + {y^2}{\sin ^2}\theta - xy\sin 2\theta = {d^2}
Adding both the equations we get,
x2cos⁡2θ+y2sin⁡2θ+xysin⁡2θ+x2cos⁡2θ+y2sin⁡2θ−xysin⁡2θ=c2+d2{x^2}{\cos ^2}\theta + {y^2}{\sin ^2}\theta + xy\sin 2\theta + {x^2}{\cos ^2}\theta + {y^2}{\sin ^2}\theta - xy\sin 2\theta = {c^2} + {d^2}
By using the identity written on the formula we get,
x2cos⁡2θ+y2(1−cos⁡2θ)+xysin⁡2θ+x2cos⁡2θ+y2(1−cos⁡2θ)−xysin⁡2θ=c2+d2{x^2}{\cos ^2}\theta + {y^2}\left( {1 - {{\cos }^2}\theta } \right) + xy\sin 2\theta + {x^2}{\cos ^2}\theta + {y^2}\left( {1 - {{\cos }^2}\theta } \right) - xy\sin 2\theta = {c^2} + {d^2}
The coefficients of “ sin⁡2θ\sin 2\theta “ terms occurs twice and their sum is zero. So it can be removed.
⇒2x2cos⁡2θ+2y2(1−cos⁡2θ)=c2+d2\Rightarrow 2{x^2}{\cos ^2}\theta + 2{y^2}\left( {1 - {{\cos }^2}\theta } \right) = {c^2} + {d^2}
⇒2x2(c+d2x)2+2y2[1−(c+d2x)2]=c2+d2\Rightarrow 2{x^2}{\left( {\dfrac{{c + d}}{{2x}}} \right)^2} + 2{y^2}\left[ {1 - {{\left( {\dfrac{{c + d}}{{2x}}} \right)}^2}} \right] = {c^2} + {d^2}
⇒2x2×(c+d)24x2+2y2[1−(c+d)24x2]=c2+d2\Rightarrow 2{x^2} \times \dfrac{{{{\left( {c + d} \right)}^2}}}{{4{x^2}}} + 2{y^2}\left[ {1 - \dfrac{{{{\left( {c + d} \right)}^2}}}{{4{x^2}}}} \right] = {c^2} + {d^2}
⇒(c+d)22+y2[4x2−(c+d)22x2]=c2+d2\Rightarrow \dfrac{{{{\left( {c + d} \right)}^2}}}{2} + {y^2}\left[ {\dfrac{{4{x^2} - {{\left( {c + d} \right)}^2}}}{{2{x^2}}}} \right] = {c^2} + {d^2}
The above is the desired equation after eliminating θ\theta .
From (1) we have,
cos⁡θ=c+d2x⇒θ=cos⁡−1(c+d2x)\cos \theta = \dfrac{{c + d}}{{2x}} \Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{{c + d}}{{2x}}} \right)
And this is the eliminant of the given equation.
So, the correct answer is “θ=cos⁡−1(c+d2x)\theta = {\cos ^{ - 1}}\left( {\dfrac{{c + d}}{{2x}}} \right) ”.

Note : While dealing with two equations and eliminating some variables always look into the functions present there. Like in our case it involved trigonometric functions sine and cosine and we used the identity that the sum of squares of sine and cosine is unity. So we squared the equations to use this thing. You may need to use the logarithmic or exponential function in some other cases.