If cos x = cos α cos β then prove that tan(\(\frac{x+a}{2}\)).tan�

1.	If cos x = cos α cos β then prove that tan(\(\frac{x+a}{2}\)).tan (\(\frac{x-a}{2}\) ) = \(\frac{tan^2\beta}{2}\)
Answer» Given, cosx = cosα cosβ ∴ cos β = \(\frac{cosx}{cos\alpha}\) ...(i) We know that, \(tan^2\frac{θ}{2}\) = \(\frac{1-cosθ}{1+cosθ}\) R. H. S = \(tan^2\frac{θ}{2}\) = \(\frac{1-cos\beta}{1+cos\beta}\) ⇒ \(\frac{1-\frac{cos\alpha}{cosx}}{1+\frac{cosx}{cos\alpha}}\) ⇒\(\frac{cos\alpha-cosx}{cos\alpha +cosx}\) = \(\frac{2sin(\frac{a+x}{2})sin(\frac{x-a}{2})}{2cos(\frac{x+a}{2})sin(\frac{x-a}{2}))}\) = \(tan(\frac{x+a}{2}).tan(\frac{x-a}{2})\) [ ∵ cosC − cosD = −2 sin (\(\frac{C+D}{2}\)) sin (\(\frac{C-D}{2}\)) and cosC + cosD = 2 cos (\(\frac{C+D}{2}\)) cos (\(\frac{C-D}{2}\)) ] = L. H. S Hence Proved

If cos x = cos α cos β then prove that tan(\(\frac{x+a}{2}\)).tan (\(\frac{x-a}{2}\) ) = \(\frac{tan^2\beta}{2}\)

Answer»

Given, cosx = cosα cosβ

∴ cos β = \(\frac{cosx}{cos\alpha}\) ...(i)

We know that, \(tan^2\frac{θ}{2}\) = \(\frac{1-cosθ}{1+cosθ}\)

R. H. S = \(tan^2\frac{θ}{2}\) = \(\frac{1-cos\beta}{1+cos\beta}\)

⇒ \(\frac{1-\frac{cos\alpha}{cosx}}{1+\frac{cosx}{cos\alpha}}\)

⇒\(\frac{cos\alpha-cosx}{cos\alpha +cosx}\)

= \(\frac{2sin(\frac{a+x}{2})sin(\frac{x-a}{2})}{2cos(\frac{x+a}{2})sin(\frac{x-a}{2}))}\)

= \(tan(\frac{x+a}{2}).tan(\frac{x-a}{2})\)

[ ∵ cosC − cosD = −2 sin (\(\frac{C+D}{2}\)) sin (\(\frac{C-D}{2}\))

and cosC + cosD = 2 cos (\(\frac{C+D}{2}\)) cos (\(\frac{C-D}{2}\)) ]

= L. H. S Hence Proved