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Question Number 193423 by mustafazaheen last updated on 13/Jun/23

$$\mathrm{when}\tan \frac{\theta }{2}=\frac{1}{\mathrm{a}}$$$$\mathrm{then}\mathrm{find}\cos \theta =?\mathrm{from}\mathrm{the}\mathrm{a}$$

Answered by deleteduser1 last updated on 13/Jun/23

$$tan(\frac{\theta }{2}+\frac{\theta }{2})=\frac{2tan\left(\frac{\theta }{2}\right)}{1-{tan}^2\left(\frac{\theta }{2}\right)}\Rightarrow tan\left(\theta \right)=\frac{2a}{a^2-1}$$$${sec}^2\left(\theta \right)=1+{tan}^2\theta =1+\frac{4a^2}{{(a^2-1)}^2}=\frac{{(a^2-1)}^2+4a^2}{{(a^2-1)}^2}$$$$\Rightarrow cos\theta =\frac{a^2-1}{\sqrt{{(a^2+1)}^2}}=\frac{a^2-1}{a^2+1}=1-\frac{2}{a^2+1}$$

Commented by mustafazaheen last updated on 13/Jun/23

$$\underbrace{T}$$

Answered by Subhi last updated on 13/Jun/23

$$cos\left(\theta \right)=1-2{sin}^2\left(\frac{\theta }{2}\right)=2{cos}^2\left(\frac{\theta }{2}\right)-1$$$$sin\left(\frac{\theta }{2}\right)=\sqrt{\frac{1-cos\left(\theta \right)}{2}}$$$$cos\left(\frac{\theta }{2}\right)=\sqrt{\frac{1+cos\left(\theta \right)}{2}}$$$$\frac{sin\left(\frac{\theta }{2}\right)}{cos\left(\frac{\theta }{2}\right)}=tan\left(\frac{\theta }{2}\right)=\frac{1}{a}=\sqrt{\frac{1-cos\left(\theta \right)}{1+cos\left(\theta \right)}}$$$$a^2(1-cos\left(\theta \right))=1+cos\left(\theta \right)\Rrightarrow a^2-1=(1+a^2)cos\left(\theta \right)$$$$cos\left(\theta \right)=\frac{a^2-1}{a^2+1}$$

Commented by mustafazaheen last updated on 13/Jun/23

$$\cancel{⋚}$$

Commented by York12 last updated on 16/Jun/23

$$\cancel{⋚}$$

Answered by BaliramKumar last updated on 13/Jun/23

$$\cos \theta =\frac{1-{\tan }^2\left(\frac{\theta }{2}\right)}{1+{\tan }^2\left(\frac{\theta }{2}\right)}=\frac{1-{\left(\frac{1}{\mathrm{a}}\right)}^2}{1+{\left(\frac{1}{\mathrm{a}}\right)}^2}=\frac{1-\frac{1}{{\mathrm{a}}^2}}{1+\frac{1}{{\mathrm{a}}^2}}$$$$\cos \theta =\frac{{\mathrm{a}}^2-1}{{\mathrm{a}}^2+1}$$$$\mathrm{pythagorean}\mathrm{triplet}\Rightarrow \underset{\mathrm{P}}{\underbrace{2\mathrm{a}}},\underset{\mathrm{B}}{\underbrace{({\mathrm{a}}^2-1)}},\underset{\mathrm{H}}{\underbrace{({\mathrm{a}}^2+1)}}$$$$\tan \theta =\frac{2\mathrm{a}}{{\mathrm{a}}^2-1}$$

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