Question

If a = x cos $\theta$ + y sin $\theta$ and b = x sin $\theta$ - y cos $\theta$ , then obtain a relation in x and y by eliminating ' $\theta$ '.

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Answer 1

Given :

If a = x cos $\theta$ + y sin $\theta$ and b = x sin $\theta$ - y cos $\theta$ , then obtain a relation in x and y by eliminating ' $\theta$ '.

Solution :

$\normalsize\sf\ Given \: that \: a \: = \: x \: cos \: \theta \: + \: y \: sin \: \theta$

$\large\sf\ squaring \: on \: both \: sides, \: we \: get$

$\large\sf\ {a}^{2} \: = \: {( \: x \: cos \: \theta \: + \: y \: sin \: \theta \: )}^{2}$

$\large\sf\ \: \: \: \: \: \: = \: {x}^{2} \: {cos}^{2} \: \theta \: + \: {y}^{2} \: {sin}^{2} \: \theta \: + \: 2.x \: cos \: \theta. \: y \: sin \: \theta \: \big( \because \: ( \: a \: + \: b \: ) {}^{2} \: = \: {a}^{2} \: + \: {b}^{2} \: + \: 2ab \: \big)$

$\large\sf\ {a}^{2} \: = \: {x}^{2} \: {cos}^{2} \: \theta \: + \: {y}^{2} \: {sin}^{2} \: \theta \: + \: 2xy \: \: sin \: \theta \: \: cos \: \theta \: \rule{30mm}{1pt} \: \: {\boxed{\large{\sf{\red{1}}}}}$

$\large\sf\ b \: \: \: = \: x \: sin \: \theta \: - \: y \: cos \: \theta$

$\large\sf\ squaring \: on \: both \: sides, \: we \: get$

$\large\sf\ {b}^{2} \: \: = \: ( \: x \: sin \: \theta \: - \: y \: cos \: \theta \: ) {}^{2}$

$\large\sf\ \: \: \: \: \: \: \: \: = \: {x}^{2} \: {sin}^{2} \: \theta \: + \: {y}^{2} \: {cos}^{2} \: \theta \: - \: 2.x \: sin \: \theta. \: y \: cos \: \theta \: \big( \because \: ( \: a \: - \: b \: ) {}^{2} \: = \: {a}^{2} \: + \: {b}^{2} \: - \: 2ab \: \big)$

$\large\sf\ {b}^{2} \: \: \: = \: {x}^{2} \: {sin}^{2} \: \theta \: + \: {y}^{2} \: {cos}^{2} \: \theta \: - \: 2xy \: sin \: \theta \: cos \: \theta \: \rule{30mm}{1pt} \: \: {\boxed{\large{\sf{\red{2}}}}}$

Now, Let's add Eq 1 + 2

$\large\sf\ {a}^{2} \: \: \: \: = \: {x}^{2} \: {cos}^{2} \: \theta \: + \: {y}^{2} \: {sin}^{2} \: \theta \: + \: 2xy \: sin \: \theta \: cos \: \theta$

$\large\sf\ {b}^{2} \: \: \: \: = \: {x}^{2} \: {sin}^{2} \: \theta \: + \: {y}^{2} \: {cos}^{2} \: \theta \: - \: 2xy \: sin \: \theta \: cos \: \theta$

$\large\sf\ {a}^{2} \: + \: {b}^{2} \: = \: {x}^{2} \: ( \: {cos}^{2} \: \theta \: + \: {sin}^{2} \: \theta \: ) \: + \: {y}^{2} \: ( \: {sin}^{2} \: \theta \: + \: {cos}^{2} \: \theta \: )$

$\large\sf\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: {x}^{2} \: (1) \: + \: {y}^{2} \: (1) \: = \: {x}^{2} \: + \: {y}^{2}$

$\tiny$

${\large{\boxed{\sf{\red{ {a}^{2} \: + \: {b}^{2} \: = \: {x}^{2} \: + \: {y}^{2}}}}}}$

$\rule{150mm}{1pt}$

Know More :

Trigomentry Identidites :

$\large\sf\ {sin}^{2} \: \theta \: + \: {cos}^{2} \: \theta \: = \: 1$

$\large\sf\ {sec}^{2} \: \theta \: - \: {tan}^{2} \: \theta \: = \: 1$

$\large\sf\ {cosec}^{2} \: \theta \: - \: {cot}^{2} \: \theta \: = \: 1$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & N.D \\ \\ \rm cosec A & N.D & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & N.D \\ \\ \rm cot A & N.D & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$