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

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

Commented by benjo_mathlover last updated on 10/Jan/21

$$x=\frac{\cos {}^2\frac{q}{2}+\sin {}^2\frac{q}{2}+2\sin \frac{q}{2}\cos \frac{q}{2}}{\cos {}^2\frac{q}{2}-\sin {}^2\frac{q}{2}}=\frac{\cos \frac{q}{2}+\sin \frac{q}{2}}{\cos \frac{q}{2}-\sin \frac{q}{2}}$$$$x^2=\frac{1+\sin q}{1-\sin q}then1+x^2=\frac{1+\sin q}{1-\sin q}+1=\frac{2}{1-\sin q}$$$$N\mathrm{ow}{\mathrm{x}}^2+(1+{\mathrm{x}}^2)\sin \mathrm{q}=\frac{1+\sin \mathrm{q}}{1-\sin \mathrm{q}}+\frac{2\sin \mathrm{q}}{1-\sin \mathrm{q}}=\frac{1+3\sin \mathrm{q}}{1-\sin \mathrm{q}}$$

Answered by bramlexs22 last updated on 10/Jan/21

$$\iff \mathrm{x}=\frac{1+\sin \mathrm{q}}{\cos \mathrm{q}};{\mathrm{x}}^2=\frac{1+2\sin \mathrm{q}+\sin {}^2\mathrm{q}}{\cos {}^2\mathrm{q}}$$$$\iff 1+{\mathrm{x}}^2=\frac{\cos {}^2\mathrm{q}+1+2\sin \mathrm{q}+\sin {}^2\mathrm{q}}{\cos {}^2\mathrm{q}}$$$$\iff \frac{2+2\sin \mathrm{q}}{\cos {}^2\mathrm{q}}\mathrm{then}{\mathrm{x}}^2+(1+{\mathrm{x}}^2)\sin \mathrm{q}=$$$$\frac{1+2\sin \mathrm{q}+\sin {}^2\mathrm{q}}{\cos {}^2\mathrm{q}}+\left(\frac{2+2\sin \mathrm{q}}{\cos {}^2\mathrm{q}}\right).\sin \mathrm{q}=$$$$\frac{1+4\sin \mathrm{q}+3\sin {}^2\mathrm{q}}{\cos {}^2\mathrm{q}}=\frac{4\sin {}^2\mathrm{q}+4\sin \mathrm{q}+\cos {}^2\mathrm{q}}{\cos {}^2\mathrm{q}}$$$$=1+\frac{4\sin \mathrm{q}(\sin \mathrm{q}+1)}{1-\sin {}^2\mathrm{q}}$$$$=1+\frac{4\sin \mathrm{q}}{1-\sin \mathrm{q}}=1+\frac{4}{\frac{1}{\sin \mathrm{q}}-1}$$$$$$

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