Math Doubts

Find the value of $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}$ if $\dfrac{x\cos{\theta}}{a}+\dfrac{y\sin{\theta}}{b} = 1$ and $\dfrac{-x\sin{\theta}}{a}+\dfrac{y\cos{\theta}}{b} = 1$

$x$ and $y$ are algebraic variables, $a$ and $b$ are constants and $\theta$ is an angle of right triangle. The elements formed two algebraic trigonometric equations.

$(1) \,\,\,\,\,\,$ $\dfrac{x\cos{\theta}}{a}+\dfrac{y\sin{\theta}}{b} = 1$

$(2) \,\,\,\,\,\,$ $\dfrac{-x\sin{\theta}}{a}+\dfrac{y\cos{\theta}}{b} = 1$

Each algebraic trigonometric equation has two terms left side and its equivalent value in right side. The two term left side expressions represent sum of terms and difference of terms.

The value of $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}$ is required to evaluate by considering both algebraic trigonometric equations in this problem.

Square both sides of first equation

Square both sides of the first equation and then expand the square of the sum of the two terms.

${\Bigg(\dfrac{x\cos{\theta}}{a}+\dfrac{y\sin{\theta}}{b}\Bigg)}^2$ $\,=\,$ ${(1)}^2$

$\implies$ ${\Bigg(\dfrac{x\cos{\theta}}{a}\Bigg)}^2$ $+$ ${\Bigg(\dfrac{y\sin{\theta}}{b}\Bigg)}^2$ $+$ $2 \times \dfrac{x\cos{\theta}}{a} \times \dfrac{y\sin{\theta}}{b}$ $\,=\,$ $1$

$\implies$ $\dfrac{x^2\cos^2{\theta}}{a^2}$ $+$ $\dfrac{y^2\sin^2{\theta}}{b^2}$ $+$ $\dfrac{2 \times x\cos{\theta} \times y\sin{\theta}}{a \times b}$ $\,=\,$ $1$

$\implies$ $\dfrac{x^2\cos^2{\theta}}{a^2}$ $+$ $\dfrac{y^2\sin^2{\theta}}{b^2}$ $+$ $\dfrac{2xy\cos{\theta}\sin{\theta}}{ab}$ $\,=\,$ $1$

Square both sides of second equation

Similarly, square both sides of the second equation and expand it.

${\Bigg(\dfrac{-x\sin{\theta}}{a}+\dfrac{y\cos{\theta}}{b}\Bigg)}^2$ $\,=\,$ ${(1)}^2$

$\implies$ ${\Bigg(\dfrac{y\cos{\theta}}{b} -\dfrac{x\sin{\theta}}{a}\Bigg)}^2$ $\,=\,$ $1$

The left hand side expression is square of subtraction of two terms and expand it.

$\implies$ ${\Bigg(\dfrac{y\cos{\theta}}{b}\Bigg)}^2$ $+$ ${\Bigg(\dfrac{x\sin{\theta}}{a}\Bigg)}^2$ $-$ $2 \times \dfrac{y\cos{\theta}}{b} \times \dfrac{x\sin{\theta}}{a}$ $\,=\,$ $1$

$\implies$ $\dfrac{y^2\cos^2{\theta}}{b^2}$ $+$ $\dfrac{x^2\sin^2{\theta}}{a^2}$ $-$ $\dfrac{2 \times y\cos{\theta} \times x\sin{\theta}}{b \times a}$ $\,=\,$ $1$

$\implies$ $\dfrac{x^2\sin^2{\theta}}{a^2}$ $+$ $\dfrac{y^2\cos^2{\theta}}{b^2}$ $-$ $\dfrac{2yx\cos{\theta}\sin{\theta}}{ba}$ $\,=\,$ $1$

$\implies$ $\dfrac{y^2\cos^2{\theta}}{b^2}$ $+$ $\dfrac{x^2\sin^2{\theta}}{a^2}$ $-$ $\dfrac{2xy\sin{\theta}\cos{\theta}}{ab}$ $\,=\,$ $1$

Add both equations

The two algebraic trigonometric equations are expanded in square form in the first and second steps.

$(1) \,\,\,\,\,\,$ $\dfrac{x^2\cos^2{\theta}}{a^2}$ $+$ $\dfrac{y^2\sin^2{\theta}}{b^2}$ $+$ $\dfrac{2xy\cos{\theta}\sin{\theta}}{ab}$ $\,=\,$ $1$

$(2) \,\,\,\,\,\,$ $\dfrac{x^2\sin^2{\theta}}{a^2}$ $+$ $\dfrac{y^2\cos^2{\theta}}{b^2}$ $-$ $\dfrac{2xy\cos{\theta}\sin{\theta}}{ab}$ $\,=\,$ $1$

Now, add both equations to eliminate third terms of the left hand expressions from both equations.

$\dfrac{x^2\cos^2{\theta}}{a^2}$ $+$ $\dfrac{y^2\sin^2{\theta}}{b^2}$ $+$ $\dfrac{2xy\cos{\theta}\sin{\theta}}{ab}$ $+$ $\dfrac{x^2\sin^2{\theta}}{a^2}$ $+$ $\dfrac{y^2\cos^2{\theta}}{b^2}$ $-$ $\dfrac{2xy\cos{\theta}\sin{\theta}}{ab}$ $\,=\,$ $1+1$

$\implies$ $\dfrac{x^2\cos^2{\theta}}{a^2}$ $+$ $\dfrac{x^2\sin^2{\theta}}{a^2}$ $+$ $\dfrac{y^2\sin^2{\theta}}{b^2}$ $+$ $\dfrac{y^2\cos^2{\theta}}{b^2}$ $+$ $\dfrac{2xy\cos{\theta}\sin{\theta}}{ab}$ $-$ $\dfrac{2xy\cos{\theta}\sin{\theta}}{ab}$ $\,=\,$ $2$

$\implies$ $\dfrac{x^2\cos^2{\theta}}{a^2}$ $+$ $\dfrac{x^2\sin^2{\theta}}{a^2}$ $+$ $\dfrac{y^2\sin^2{\theta}}{b^2}$ $+$ $\dfrac{y^2\cos^2{\theta}}{b^2}$ $+$ $\require{cancel} \cancel{\dfrac{2xy\cos{\theta}\sin{\theta}}{ab}}$ $-$ $\require{cancel} \cancel{\dfrac{2xy\cos{\theta}\sin{\theta}}{ab}}$ $\,=\,$ $2$

$\implies$ $\dfrac{x^2\cos^2{\theta}}{a^2}$ $+$ $\dfrac{x^2\sin^2{\theta}}{a^2}$ $+$ $\dfrac{y^2\sin^2{\theta}}{b^2}$ $+$ $\dfrac{y^2\cos^2{\theta}}{b^2}$ $\,=\,$ $2$

$\implies$ $\dfrac{x^2}{a^2}{(\cos^2{\theta}+\sin^2{\theta})}$ $+$ $\dfrac{y^2}{b^2}{(\cos^2{\theta}+\sin^2{\theta})}$ $\,=\,$ $2$

According to Pythagorean identity of sin and cos functions, the squares of sum of sin and cos functions at an angle is always one.

$\implies$ $\dfrac{x^2}{a^2}{(1)}$ $+$ $\dfrac{y^2}{b^2}{(1)}$ $\,=\,$ $2$

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{x^2}{a^2}$ $+$ $\dfrac{y^2}{b^2}$ $\,=\,$ $2$



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