- #1
Ganesh Ujwal
- 56
- 0
Originally posted in a technical section, so missing the template
Mod note: Fixed the LaTeX.
##a=sinθ+sinϕ##
##b=tanθ+tanϕ##
##c=secθ+secϕ##Show that,
##8bc=a[4b^2 + (b^2-c^2)^2]##
I tried to solve this for hours and have gotten no-where. Here's what I've got so far :
##a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2}) ##
## b = \frac{2\sin(\theta+\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}##
##c=\frac{2(\cos\theta+\cos\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}##
##a^2 = \frac{\sin^2(\theta+\phi)[\cos(\theta+\phi)+1]}{\cos(\theta+\phi)+1}##
##\\cos(\theta-\phi)=\frac{ca}{b}-1##
##sin^2(\frac{\theta+\phi}{2})=\frac{2a^2b}{4(ca+b)}##
##a=sinθ+sinϕ##
##b=tanθ+tanϕ##
##c=secθ+secϕ##Show that,
##8bc=a[4b^2 + (b^2-c^2)^2]##
I tried to solve this for hours and have gotten no-where. Here's what I've got so far :
##a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2}) ##
## b = \frac{2\sin(\theta+\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}##
##c=\frac{2(\cos\theta+\cos\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}##
##a^2 = \frac{\sin^2(\theta+\phi)[\cos(\theta+\phi)+1]}{\cos(\theta+\phi)+1}##
##\\cos(\theta-\phi)=\frac{ca}{b}-1##
##sin^2(\frac{\theta+\phi}{2})=\frac{2a^2b}{4(ca+b)}##
Last edited: