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Question Number 112672 by bemath last updated on 09/Sep/20

$$\frac{\sin \mathrm{A}-\cos \mathrm{A}+1}{\sin \mathrm{A}+\cos \mathrm{A}-1}?$$

Commented by som(math1967) last updated on 09/Sep/20

$$\frac{1+\mathrm{sinA}}{\mathrm{cosA}}$$

Commented by bemath last updated on 09/Sep/20

$$\mathrm{thank}\mathrm{you}\mathrm{all}.\mathrm{all}\mathrm{answer}\mathrm{is}\mathrm{correct}$$

Answered by bobhans last updated on 09/Sep/20

$$\left(⧫\right)\frac{(\sin \mathrm{A}-\cos \mathrm{A})+1}{(\sin \mathrm{A}+\cos \mathrm{A})-1}\times \frac{(\sin \mathrm{A}+\cos \mathrm{A})+1}{(\sin \mathrm{A}+\cos \mathrm{A})+1}=$$$$\frac{\sin {}^2\mathrm{A}-\cos {}^2\mathrm{A}+\sin \mathrm{A}-\cos \mathrm{A}+\sin \mathrm{A}+\cos \mathrm{A}+1}{1+2\sin \mathrm{Acos}\mathrm{A}-1}=$$$$\frac{\sin {}^2\mathrm{A}-(1-\sin {}^2\mathrm{A})+2\sin \mathrm{A}+1}{2\sin \mathrm{Acos}\mathrm{A}}=$$$$\frac{2\sin {}^2\mathrm{A}+2\sin \mathrm{A}}{2\sin \mathrm{Acos}\mathrm{A}}=\frac{2\sin \mathrm{A}(\sin \mathrm{A}+1)}{2\sin \mathrm{Acos}\mathrm{A}}$$$$=\frac{\sin \mathrm{A}+1}{\cos \mathrm{A}}\mathrm{or}\frac{\sin \mathrm{A}+1}{\cos \mathrm{A}}\times \frac{\sin \mathrm{A}-1}{\sin \mathrm{A}-1}$$$$=\frac{\sin {}^2\mathrm{A}-1}{\cos \mathrm{A}(\sin \mathrm{A}-1)}=\frac{\cos \mathrm{A}}{1-\sin \mathrm{A}}$$

Commented by som(math1967) last updated on 09/Sep/20

$$\frac{\mathrm{cosA}(1+\mathrm{sinA})}{{\cos }^2\mathrm{A}}=\frac{1+\mathrm{sinA}}{\mathrm{cosA}}$$

Commented by bemath last updated on 09/Sep/20

$$\mathrm{same}\mathrm{sir}.\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}}=\frac{\cos \mathrm{A}}{1-\sin \mathrm{A}}$$

Commented by som(math1967) last updated on 09/Sep/20

$$\mathrm{yes}\mathrm{sir}$$

Answered by Dwaipayan Shikari last updated on 09/Sep/20

$$\frac{2sin\frac{A}{2}cos\frac{A}{2}+2{sin}^2\frac{A}{2}}{2sin\frac{A}{2}(cos\frac{A}{2}-sin\frac{A}{2})}=\frac{{(cos\frac{A}{2}+sin\frac{A}{2})}^2}{({cos}^2\frac{A}{2}-{sin}^2\frac{A}{2})}=\frac{1+sinA}{cosA}=tanA+secA$$

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