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Question Number 107364 by bemath last updated on 10/Aug/20

$$⊚\mathcal{BEMATH}⊚$$$$If\tan x+\sec x=b\Rightarrow \cos x=?$$

Answered by bobhans last updated on 10/Aug/20

$$\Rrightarrow \mathbb{BOB}\mathcal{HANS}\Lleftarrow$$$$\frac{\tan \mathrm{x}-\sec \mathrm{x}}{\tan \mathrm{x}-\sec \mathrm{x}}\times \tan \mathrm{x}+\sec \mathrm{x}=\mathrm{♭}$$$$\frac{\tan {}^2\mathrm{x}-\sec {}^2\mathrm{x}}{\tan \mathrm{x}-\sec \mathrm{x}}=\mathrm{♭}\iff \frac{-1}{\tan \mathrm{x}-\sec \mathrm{x}}=\mathrm{♭}$$$$\tan \mathrm{x}-\sec \mathrm{x}=-\frac{1}{\mathrm{♭}}$$$$\tan \mathrm{x}+\sec \mathrm{x}=\mathrm{♭}$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}-$$$$-2\sec \mathrm{x}=-\frac{1}{\mathrm{♭}}-\mathrm{♭}\Rightarrow \sec \mathrm{x}=\frac{{\mathrm{♭}}^2+1}{2\mathrm{♭}}$$$$\iff \therefore \cos \mathrm{x}=\frac{2\mathrm{♭}}{{\mathrm{♭}}^2+1}$$

Answered by som(math1967) last updated on 10/Aug/20

$${\sec }^2\mathrm{x}-{\tan }^2\mathrm{x}=1$$$$\Rightarrow \mathrm{secx}-\mathrm{tanx}=\frac{1}{\mathrm{b}}[\because \mathrm{secx}+\mathrm{tanx}=\mathrm{b}]$$$$\therefore 2\mathrm{secx}=\mathrm{b}+\frac{1}{\mathrm{b}}$$$$\mathrm{secx}=\frac{{\mathrm{b}}^2+1}{2\mathrm{b}}\therefore \mathrm{cosx}=\frac{2\mathrm{b}}{{\mathrm{b}}^2+1}\mathrm{ans}$$

Answered by Dwaipayan Shikari last updated on 10/Aug/20

$${\sec }^2\mathrm{x}-{\tan }^2\mathrm{x}=1$$$$\mathrm{b}(\mathrm{tanx}-\mathrm{secx})=-1$$$$\mathrm{tanx}+\mathrm{secx}=\mathrm{b}$$$$\mathrm{tanx}-\mathrm{secx}=-\frac{1}{\mathrm{b}}$$$$2\mathrm{secx}=\mathrm{b}+\frac{1}{\mathrm{b}}$$$$\mathrm{cosx}=\frac{2\mathrm{b}}{{\mathrm{b}}^2+1}$$

Answered by Dwaipayan Shikari last updated on 10/Aug/20

$${\tan }^2\mathrm{x}+{\sec }^2\mathrm{x}+2\mathrm{tanxsecx}={\mathrm{b}}^2$$$${2\tan }^2\mathrm{x}+1+2\mathrm{tanxsecx}={\mathrm{b}}^2$$$$2\mathrm{tanx}(\mathrm{tanx}+\mathrm{secx})={\mathrm{b}}^2-1$$$$\mathrm{tanx}=\frac{{\mathrm{b}}^2-1}{2\mathrm{b}}$$$$\frac{{\sin }^2\mathrm{x}}{{\cos }^2\mathrm{x}}+1=\frac{{({\mathrm{b}}^2-1)}^2}{{4\mathrm{b}}^2}+1$$$$\frac{1}{{\cos }^2\mathrm{x}}=\frac{{({\mathrm{b}}^2+1)}^2}{{4\mathrm{b}}^2}$$$$\mathrm{cosx}=\frac{2\mathrm{b}}{{\mathrm{b}}^2+1}$$

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