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Question Number 116630 by bobhans last updated on 05/Oct/20

$$\mathrm{Given}\mathrm{cosec}\mathrm{x}+\cot \mathrm{x}=\mathrm{p},\mathrm{find}\mathrm{the}$$$$\mathrm{value}\mathrm{of}\mathrm{cosec}\mathrm{x}=?$$

Answered by TANMAY PANACEA last updated on 05/Oct/20

$$1+{cot}^{2}x={cosec}^{2}x$$$$1=(cosecx+cotx)(cosecx-cotx)$$$$cosecx+cotx=p$$$$cosecx-cotx=\frac{1}{p}$$$$2cosecx=(p+\frac{1}{p})$$$$cosecx=\frac{1}{2}(p+\frac{1}{p})$$

Answered by MJS_new last updated on 05/Oct/20

$$\frac{1}{\sin x}+\frac{\cos x}{\sin x}=p$$$$\frac{1+\sqrt{1-{\sin }^{2}x}}{\sin x}=p$$$$\Rightarrow \sin x=\frac{2p}{p^2+1}\Rightarrow \mathrm{cosec}x=\frac{p^2+1}{2p}$$

Answered by bemath last updated on 05/Oct/20

$$\frac{1}{\sin \mathrm{x}}+\frac{\cos \mathrm{x}}{\sin \mathrm{x}}=\mathrm{p}\Rightarrow 1+\cos \mathrm{x}=\mathrm{psin}\mathrm{x}$$$$\Rightarrow \sqrt{1-\sin {}^2\mathrm{x}}=\mathrm{psin}\mathrm{x}-1$$$$\Rightarrow 1-\sin {}^2\mathrm{x}={\mathrm{p}}^2\sin {}^2\mathrm{x}-2\mathrm{psin}\mathrm{x}+1$$$$(1+{\mathrm{p}}^2)\sin {}^2\mathrm{x}=2\mathrm{psin}\mathrm{x}$$$$\sin \mathrm{x}\ne 0\Rightarrow (1+{\mathrm{p}}^2)\sin \mathrm{x}=2\mathrm{p}$$$$\Rightarrow \sin \mathrm{x}=\frac{2\mathrm{p}}{1+{\mathrm{p}}^2}\mathrm{and}\mathrm{cosec}\mathrm{x}=\frac{1+{\mathrm{p}}^2}{2\mathrm{p}}$$

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