Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 54626 by Aditya789 last updated on 08/Feb/19

$$\frac{1+\mathrm{sinx}+\mathrm{cox}}{1+\mathrm{sinx}-\mathrm{cosx}}=\frac{1-\mathrm{cosx}}{1+\mathrm{cosx}}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 08/Feb/19

$$LHS$$$$\frac{2{cos}^2\frac{x}{2}+2sin\frac{x}{2}cos\frac{x}{2}}{2{sin}^2\frac{x}{2}+2sin\frac{x}{2}cos\frac{x}{2}}$$$$\frac{2cos\frac{x}{2}(cos\frac{x}{2}+sin\frac{x}{2})}{2sin\frac{x}{2}(sin\frac{x}{2}+cos\frac{x}{2})}=cot\frac{x}{2}$$$$RHS$$$$\frac{2{sin}^2\frac{x}{2}}{2{cos}^2\frac{x}{2}}={tan}^2\frac{x}{2}$$$$soplsrecheckthequestion...$$

Answered by MJS last updated on 08/Feb/19

$$1+\cos x\ne 0\Rightarrow x\ne 2z\pi -\pi ;z\in \mathbb{Z}$$$$1+\sin x-\cos x\ne 0\Rightarrow x\ne 2z\pi \wedge x\ne 2z\pi -\frac{\pi }{2};z\in \mathbb{Z}$$$$\frac{1+s+c}{1+s-c}=\frac{1-c}{1+c}$$$$(1+s+c)(1+c)=(1-c)(1+s-c)$$$$2sc+4c=0$$$$c(s+2)=0$$$$\cos x(2+\sin x)=0$$$$\Rightarrow \cos x=0\Rightarrow x=z\pi -\frac{\pi }{2}\mathrm{but}x\ne 2z\pi -\frac{\pi }{2}$$$$\Rightarrow x=(2z-1)\pi -\frac{\pi }{2}=2z\pi -\frac{3\pi }{2}=\frac{\pi }{2}+2z\pi ;z\in \mathbb{Z}$$$$x=\frac{\pi }{2}+2z\pi ;z\in \mathbb{Z}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com