If-x-cos-q-sin-q-1-then-x-2-1-x-2-sin-q-

Question Number 128808 by liberty last updated on 10/Jan/21

If x \cos q - \sin q = 1 then x^{2} + (1 + x^{2}) \sin q = ?

Commented by benjo_mathlover last updated on 10/Jan/21

x = \frac{\cos^{2} \frac{q}{2} + \sin^{2} \frac{q}{2} + 2 \sin \frac{q}{2} \cos \frac{q}{2}}{\cos^{2} \frac{q}{2} - \sin^{2} \frac{q}{2}} = \frac{\cos \frac{q}{2} + \sin \frac{q}{2}}{\cos \frac{q}{2} - \sin \frac{q}{2}}

x^{2} = \frac{1 + \sin q}{1 - \sin q} t h e n 1 + x^{2} = \frac{1 + \sin q}{1 - \sin q} + 1 = \frac{2}{1 - \sin q}

N ow x^{2} + (1 + x^{2}) \sin q = \frac{1 + \sin q}{1 - \sin q} + \frac{2 \sin q}{1 - \sin q} = \frac{1 + 3 \sin q}{1 - \sin q}

Answered by bramlexs22 last updated on 10/Jan/21

\Leftrightarrow x = \frac{1 + \sin q}{\cos q}; x^{2} = \frac{1 + 2 \sin q + \sin^{2} q}{\cos^{2} q}

\Leftrightarrow 1 + x^{2} = \frac{\cos^{2} q + 1 + 2 \sin q + \sin^{2} q}{\cos^{2} q}

\Leftrightarrow \frac{2 + 2 \sin q}{\cos^{2} q} then x^{2} + (1 + x^{2}) \sin q =

\frac{1 + 2 \sin q + \sin^{2} q}{\cos^{2} q} + (\frac{2 + 2 \sin q}{\cos^{2} q}) . \sin q =

\frac{1 + 4 \sin q + 3 \sin^{2} q}{\cos^{2} q} = \frac{4 \sin^{2} q + 4 \sin q + \cos^{2} q}{\cos^{2} q}

= 1 + \frac{4 \sin q (\sin q + 1)}{1 - \sin^{2} q}

= 1 + \frac{4 \sin q}{1 - \sin q} = 1 + \frac{4}{\frac{1}{\sin q} - 1}