Question

Prove that $tan[2 tan^{-1} (\frac{\sqrt{5} - 1}{2})]$ = 2.

Answer 1

Please see the attachment

Answer 2

Answer:

Step-by-step explanation:

As we know that

$2\:tan^{-1}x= tan^{-1}(\frac{2x}{1-x^{2} })\\\\\\2\:tan^{-1}(\frac{\sqrt{5}-1 }{2}) = tan^{-1}(\frac{2(\frac{\sqrt{5}-1 }{2})}{1-(\frac{\sqrt{5}-1 }{2})^{2} })\\\\\\=tan^{-1}(\frac{\sqrt{5}-1 }{\frac{4-(5+1-2\sqrt{5})}{4} })\\\\\\=tan^{-1}(\frac{\sqrt{5}-1 }{\frac{4-6+2\sqrt{5})}{4} })\\\\\\=tan^{-1}(\frac{\sqrt{5}-1 }{\frac{2\sqrt{5}-2)}{4} })\\\\\\=tan^{-1}(\frac{\sqrt{5}-1 }{\frac{2(\sqrt{5}-1)}{4}})\\\\\\=tan^{-1}(\frac{4\sqrt{5}-1 }{2\sqrt{5}-1})\\\\\\=tan^{-1}(2)$

$tan[2 tan^{-1} (\frac{\sqrt{5} - 1}{2})]=tan\:tan^{-1}(2)\\\\=2$

hence proved