Question

Question 9:-
Q9. If sin 0 + cos 0 =y, and cos 0 - sin 0 = x, then
x2 + y 2 is equal to
(a) o
(b) 1/2
(c) 1
(d) 2
​

Answer 1

$y = \sin( \theta) + \cos(\theta) \\ squaring \: both \: side \\ {y}^{2} = ( \sin(\theta) + \cos(\theta) ) {}^{2} \\ = \sin {}^{2} ( \theta) + 2 \sin(\theta) \cos(\theta) + \cos {}^{2} (\theta) \\ = \{\sin {}^{2} ( \theta) + \cos {}^{2} (\theta) \} +2 \sin(\theta) \cos(\theta) \\ = 1 + 2 \sin(\theta) \cos(\theta) \\ = > {y}^{2} = 1 + 2 \sin(\theta) \cos(\theta)$

$x = \cos( \theta) - \sin( \theta) \\ squaring \: both \: side \\ {x}^{2} = (\cos( \theta) - \sin( \theta) ) {}^{2} \\ = \cos {}^{2} ( \theta) + sin {}^{2} ( \theta) -2 \sin( \theta) \cos( \theta) \\ = 1 -2 \sin( \theta) \cos( \theta) \\ = > {x}^{2} = 1 -2 \sin( \theta) \cos( \theta)$

$then \\ {x}^{2} + {y}^{2} \\ = 1 + 2 \sin( \theta) \cos( \theta) + 1 -2 \sin( \theta) \cos( \theta) \\ = 2$

$(d) \: is \: the \: correct \: answer$