Question

Express the equations in terms of x and y if x cos theta + y sin theta = a;
x sin theta - ycos theta = b​

Answer 1

$\red\bigstar$ EXPLAINATION $\red\bigstar$

$a = xcos\theta + ysin\theta$

$b = xsin\theta - ycos\theta$

$a^2 = (xcos\theta + ysin\theta)^2 = x^2cos^2\theta + y^2sin^2\theta + 2xysin\theta\cos\theta$

$b^2 = (xsin\theta - ycos\theta)^2 = x^2sin^2\theta + y^2cos^2\theta - 2xysin\theta cos\theta$

$a^2 + b^2 = x^2cos^2\theta +y^2sin^2\theta + 2xysin\theta cos\theta + x^2sin^2\theta + y^2cos\theta - 2xysin\theta cos\theta$

$"2xysin\theta cos\theta"\: "-2xysin\theta cos\theta"\: got\: cancelled$

$a^2 + b^2 = x^2(cos^2\theta + sin^2\theta) + y^2(sin^2\theta + cos^2\theta)$

$We\: know\: that\: sin^2\theta + cos^2\theta = 1$

$Therefore,$

$a^2 + b^2 = x^2 + y^2$

$\red\star$ EXTRA INFORMATION $\red\star$

i)$sin\theta = \frac{Perpendicular}{Hypotenuse}$

ii)$cos\theta = \frac{Base}{Hypotenuse}$

iii)$tan\theta = \frac{Perpendicular}{Base}$

iv)$cot\theta = \frac{Base}{Perpendicular}$

v)$sec\theta = \frac{Hypotenuse}{Base}$

vi)$cosec\theta = \frac{Hypotenuse}{Perpendicular}$

vii)$sin\theta = \frac{1}{cosec\theta}$

viii)$cos\theta = \frac{1}{sec\theta}$

ix)$tan\theta = \frac{1}{cot\theta}$

x)$sin^2\theta + cos^2\theta = 1$

xi)$1 + tan^2\theta = sec^2\theta$

xii)$1 + cot^2\theta = cosec^2\theta$