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Question Number 173894 by azadsir last updated on 20/Jul/22

$$\mathrm{If}\mathrm{secA}-\mathrm{tanA}=\mathrm{Q}\mathrm{than}\mathrm{prove}\mathrm{that},$$$$\mathrm{cosecA}=\frac{1+{\mathrm{Q}}^2}{1-{\mathrm{Q}}^2}$$

Commented by cortano1 last updated on 20/Jul/22

$$Q=\frac{1-\sin A}{\cos A}\Rightarrow Q^2\cos {}^{2}A=1+\sin {}^{2}A-2\sin A$$$$\Rightarrow \sin {}^{2}A-Q^2(1-\sin {}^{2}A)-2\sin A+1=0$$$$\Rightarrow (1+Q^2)\sin {}^{2}A-2\sin A+1-Q^2=0$$$$\Rightarrow \sin A=\frac{2\pm \sqrt{4-4(1+Q^2)(1-Q^2)}}{2(1+Q^2)}$$$$\Rightarrow \sin A=\frac{2\pm \sqrt{4-4(1-Q^4)}}{2(1+Q^2)}$$$$\Rightarrow \csc A=\frac{2(1+Q^2)}{2\pm 2Q^2}=\frac{1+Q^2}{1\pm Q^2}$$$$\Rightarrow \sin A\ne 1\Rightarrow \csc A=\frac{1+Q^2}{1-Q^2}$$

Commented by azadsir last updated on 20/Jul/22

$$\mathrm{Thank}\mathrm{you}$$

Answered by blackmamba last updated on 20/Jul/22

$$Q=\frac{1-\sin A}{\cos A}=\frac{\cos \frac{1}{2}A-\sin \frac{1}{2}A}{\cos \frac{1}{2}A+\sin \frac{1}{2}A}$$$$Q=\frac{1-\tan \frac{1}{2}A}{1+\tan \frac{1}{2}A}$$$$Q+Q\tan \frac{1}{2}A=1-\tan \frac{1}{2}A$$$$\tan \frac{1}{2}A=\frac{1-Q}{1+Q}$$$$\{\begin{array}{l}\sin \frac{1}{2}A=\frac{1-Q}{\sqrt{2+2Q^2}}\\ \cos \frac{1}{2}=\frac{1+Q}{\sqrt{2+2Q^2}}\end{array}$$$$\sin A=\frac{2(1-Q^2)}{2(1+Q^2)}$$$$\frac{1}{\sin A}=\frac{1+Q^2}{1-Q^2}$$$$$$$$$$

Commented by azadsir last updated on 20/Jul/22

$$\mathrm{Thank}\mathrm{you}$$

Commented by Tawa11 last updated on 21/Jul/22

$$\mathrm{Great}\mathrm{sirs}$$

Answered by BaliramKumar last updated on 21/Jul/22

$$secA-tanA=Q$$$$\frac{1-sinA}{cosA}=Q$$$$\frac{{(1-sinA)}^2}{{cos}^{2}A}=Q^2$$$$\frac{{(1-sinA)}^2}{1-{sin}^{2}A}=Q^2$$$$\frac{{(1-sinA)}^2}{(1-sinA)(1+sinA)}=Q^2$$$$\frac{(1-sinA)}{(1+sinA)}=Q^2$$$$\frac{(1+sinA)}{(1-sinA)}=\frac{1}{Q^2}$$$$\frac{(1+sinA)+(1-sinA)}{(1+sinA)-(1-sinA)}=\frac{1+Q^2}{1-Q^2}[\because If\frac{a}{b}=\frac{c}{d}then\frac{a+b}{a-b}=\frac{c+d}{c-d}]$$$$\frac{2}{2sinA}=\frac{1+Q^2}{1-Q^2}$$$$\frac{1}{sinA}=\frac{1+Q^2}{1-Q^2}$$$$cosecA=\frac{1+Q^2}{1-Q^2}$$$$$$

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