Question

If $x\cos \theta +y\sin \theta =z$, then what is the value of $(x\sin \theta -y\cos \theta {)}^2$?

• A. $x^2+y^2-z^2$
• B. $x^2-y^2-z^2$
• C. $x^2-y^2+z^2$
• D. $x^2+y^2+z^2$

Answer 1

The correct option is A $x^2+y^2-z^2$

$(x\sin \theta +y\cos \theta {)}^2=z^2$

$x^2{\sin }^2\theta +y^2{\cos }^2\theta +2xy\sin \theta \cos \theta =z^2$ ...(1)

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

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

$=x^2+y^2-x^2{\cos }^2\theta -y^2{\sin }^2\theta -2xy\sin \theta \cos \theta$

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

$=x^2+y^2-z^2$ ...(from 1)

So, the answer is option (A)