Question

If ${\sin }^2\theta +\cos \theta =x$, prove that ${\sin }^2\theta +{\cos }^2\theta =1$

Answer 1

Consider the given equation.

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

Since,

${\sin }^2\theta =1-{\cos }^2\theta$ ……. (2)

Therefore,

$1-{\cos }^2\theta +\cos \theta =x$

${\cos }^2\theta -\cos \theta +x-1=0$

$\cos \theta =\frac{1\pm \sqrt{1-4\times 1\times (x-1)}}{2}$

$\cos \theta =\frac{1\pm \sqrt{1-4(x-1)}}{2}$

$\cos \theta =\frac{1\pm \sqrt{1-4x+4}}{2}$

$\cos \theta =\frac{1\pm \sqrt{5-4x}}{2}$

From equation (2),

${\sin }^2\theta =1-{\left(\frac{1\pm \sqrt{5-4x}}{2}\right)}^2$

Now,

${\sin }^2\theta +{\cos }^2\theta =1-{\left(\frac{1\pm \sqrt{5-4x}}{2}\right)}^2+{\left(\frac{1\pm \sqrt{5-4x}}{2}\right)}^2$

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

Hence, this is the answer.