If x - a sin- and y -btan-then prove that t 2- x 2- b 2- y 2-1

A^2/x^2 b^2/y^2=a^2/a^2sin^2 θ b^2/b^2tan^2 θ 1/sin^2 θ 1/tan^2 θ =cosec^2 θ cot^2 θ =1=RHS ...

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Byju's Answer

Standard X

Mathematics

Relation between Trigonometric Ratios

If x = a sinθ...

Question

If $x = a s i n θ a n d y = b t a n θ$ ,then prove that $\frac{a^{2}}{x^{2}} - \frac{b^{2}}{y^{2}} = 1$

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Solution

$\frac{a^{2}}{x^{2}} - \frac{b^{2}}{y^{2}} = \frac{a^{2}}{a^{2} s i n^{2} θ} - \frac{b^{2}}{b^{2} t a n^{2} θ}$
$\frac{1}{s i n^{2} θ} - \frac{1}{t a n^{2} θ}$
$= c o s e c^{2} θ - c o t^{2} θ$
$= 1 = R H S$

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$If \frac{x}{a} c o s θ + \frac{y}{b} s i n θ = 1 and \frac{x}{a} s i n θ - \frac{y}{b} c o s θ = 1, prove that \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2$

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If x=a sinθ and y=btanθ,then prove thata2x2−b2y2=1

a2x2−b2y2=a2a2sin2θ−b2b2tan2θ 1sin2θ−1tan2θ =cosec2θ−cot2θ =1=RHS

If $x = a s i n θ a n d y = b t a n θ$ ,then prove that $\frac{a^{2}}{x^{2}} - \frac{b^{2}}{y^{2}} = 1$

$\frac{a^{2}}{x^{2}} - \frac{b^{2}}{y^{2}} = \frac{a^{2}}{a^{2} s i n^{2} θ} - \frac{b^{2}}{b^{2} t a n^{2} θ}$
$\frac{1}{s i n^{2} θ} - \frac{1}{t a n^{2} θ}$
$= c o s e c^{2} θ - c o t^{2} θ$
$= 1 = R H S$