Solveeit Logo
Login

Question

Question: If \[\sqrt {1 - {x^2}} + \sqrt {1 - {y^2}} = a(x - y)\], show that \[\dfrac{{dy}}{{dx}} = \sqrt {\df...

If 1−x2+1−y2=a(x−y)\sqrt {1 - {x^2}} + \sqrt {1 - {y^2}} = a(x - y), show that dydx=1−y21−x2\dfrac{{dy}}{{dx}} = \sqrt {\dfrac{{1 - {y^2}}}{{1 - {x^2}}}} .

Explanation

Solution

To solve this problem we have to make use of a few trigonometric identities. The formulas which we are going to used for this problem are: cos⁡2A+sin⁡2A=1{\cos ^2}A + {\sin ^2}A = 1,1−sin⁡2A=cos⁡2A1 - {\sin ^2}A = {\cos ^2}A, sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A - \sin B = 2\cos \dfrac{{A + B}}{2}\sin \dfrac{{A - B}}{2}, cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A + \cos B = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2} and sin⁡Acos⁡A=tan⁡A\dfrac{{\sin A}}{{\cos A}} = \tan A

Complete step-by-step solution:
In this given problem,
we are given as 1−x2+1−y2=a(x−y)\sqrt {1 - {x^2}} + \sqrt {1 - {y^2}} = a(x - y)
we have to solve and to show the given let us consider,
x=sin⁡Ax = \sin A and y=sin⁡By = \sin B
Hence substituting the considered value in the given question which is termed as,
1−sin⁡2A+1−sin⁡2B=a(sin⁡A−sin⁡B)\sqrt {1 - {{\sin }^2}A} + \sqrt {1 - {{\sin }^2}B} = a(\sin A - \sin B)
As we know some of the trigonometric formulas as,
cos⁡2A+sin⁡2A=1{\cos ^2}A + {\sin ^2}A = 1
1−sin⁡2A=cos⁡2A1 - {\sin ^2}A = {\cos ^2}A
Thus, 1−sin⁡2A=cos⁡2A=(cos⁡2A)12=cos⁡A\sqrt {1 - {{\sin }^2}A} = \sqrt {{{\cos }^2}A} = {({\cos ^2}A)^{\dfrac{1}{2}}} = \cos A
And 1−sin⁡2B=cos⁡2B=(cos⁡2B)12=cos⁡B\sqrt {1 - {{\sin }^2}B} = \sqrt {{{\cos }^2}B} = {({\cos ^2}B)^{\dfrac{1}{2}}} = \cos B
And with this we also know the formula as
sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A - \sin B = 2\cos \dfrac{{A + B}}{2}\sin \dfrac{{A - B}}{2}
cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A + \cos B = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}
Substituting the formulas and derived ones in the given derived termed as,
1−sin⁡2A+1−sin⁡2B=a(sin⁡A−sin⁡B)\sqrt {1 - {{\sin }^2}A} + \sqrt {1 - {{\sin }^2}B} = a(\sin A - \sin B)
⇒cos⁡A+cos⁡B=a.2cos⁡A+B2sin⁡A−B2\Rightarrow \cos A + \cos B = a.2\cos \dfrac{{A + B}}{2}\sin \dfrac{{A - B}}{2}
⇒2cos⁡A+B2cos⁡A−B2=a.2cos⁡A+B2sin⁡A−B2\Rightarrow 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2} = a.2\cos \dfrac{{A + B}}{2}\sin \dfrac{{A - B}}{2}
By comparing on both the LHS and RHS, we can find some similar terms thus cancelling out those terms as 2cos⁡A+B22\cos \dfrac{{A + B}}{2}
Then the remaining term will be as,
⇒cos⁡A−B2=asin⁡A−B2\Rightarrow \cos \dfrac{{A - B}}{2} = a\sin \dfrac{{A - B}}{2}
This can also be written as,
⇒asin⁡A−B2=cos⁡A−B2\Rightarrow a\sin \dfrac{{A - B}}{2} = \cos \dfrac{{A - B}}{2}
⇒sin⁡A−B2cos⁡A−B2=1a\Rightarrow \dfrac{{\sin \dfrac{{A - B}}{2}}}{{\cos \dfrac{{A - B}}{2}}} = \dfrac{1}{a}
As we know that sin⁡Acos⁡A=tan⁡A\dfrac{{\sin A}}{{\cos A}} = \tan A
Similarly, applying that concept in the above term we get as,
⇒tan⁡A−B2=1a\Rightarrow \tan \dfrac{{A - B}}{2} = \dfrac{1}{a}
To show the value in the given question we need to solve further as,
Taking inverse of the function,
⇒tan⁡−11a=A−B2\Rightarrow {\tan ^{ - 1}}\dfrac{1}{a} = \dfrac{{A - B}}{2}
⇒2tan⁡−11a=A−B\Rightarrow 2{\tan ^{ - 1}}\dfrac{1}{a} = A - B
From our consideration on the starting of the problem for xx and yy as
x=sin⁡Ax = \sin A and y=sin⁡By = \sin B from this finding the value of AA and BB as,
A=sin⁡−1xA = {\sin ^{ - 1}}x
B=sin⁡−1yB = {\sin ^{ - 1}}y
While substituting the derived values in the above expression we found as,
⇒2tan⁡−11a=sin⁡−1x−sin⁡−1y\Rightarrow 2{\tan ^{ - 1}}\dfrac{1}{a} = {\sin ^{ - 1}}x - {\sin ^{ - 1}}y
Differentiating the above equation with respect to xx we get as,
Similarly, here aais a constant, such that
0=11−x2−11−y2dydx0 = \dfrac{1}{{\sqrt {1 - {x^2}} }} - \dfrac{1}{{\sqrt {1 - {y^2}} }}\dfrac{{dy}}{{dx}}
Such that,
sin⁡−1x=11−x2{\sin ^{ - 1}}x = \dfrac{1}{{\sqrt {1 - {x^2}} }} and sin⁡−1y=11−y2{\sin ^{ - 1}}y = \dfrac{1}{{\sqrt {1 - {y^2}} }}
Therefore,
⇒11−y2dydx=11−x2\Rightarrow \dfrac{1}{{\sqrt {1 - {y^2}} }}\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {x^2}} }}
On cross multiply the above equation, then
⇒dydx=1−y21−x2\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {1 - {y^2}} }}{{\sqrt {1 - {x^2}} }}
From the above question using various trigonometric formula we have compute from the given as,
1−x2+1−y2=a(x−y)\sqrt {1 - {x^2}} + \sqrt {1 - {y^2}} = a(x - y)
Thus, the above solution is represented as,
dydx=1−y21−x2\dfrac{{dy}}{{dx}} = \sqrt {\dfrac{{1 - {y^2}}}{{1 - {x^2}}}}
Hence proved.

Note: Some interesting facts about trigonometry:
Fundamental identities are a set of eight trigonometric identities. Since they are based on the Pythagorean Theorem, three of them are known as Pythagorean identities.
In the 3rd century BC, applications of geometry to astronomical studies spawned the field.
Music and architecture are also synonymous with trigonometry.
Trigonometry is used by engineers to determine the angles of sound waves and how to plan spaces.