Prove it

Geometry Level pending

Which of the following is equal to

tan ⁡ θ + tan ⁡ ϕ cot ⁡ θ + cot ⁡ ϕ \large\dfrac{\tan \theta + \tan \phi}{\cot \theta + \cot \phi}

tan ⁡ θ + tan ⁡ ϕ \tan \theta+\tan \phi cot ⁡ θ cot ⁡ ϕ \cot \theta~\cot \phi tan ⁡ θ tan ⁡ ϕ \tan \theta~\tan \phi tan ⁡ θ + cot ⁡ ϕ \tan \theta+\cot \phi

A Former Brilliant Member by A Former Brilliant Member

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1 solution

tan ⁡ θ + tan ⁡ ϕ cot ⁡ θ + cot ⁡ ϕ \large\dfrac{\tan \theta + \tan \phi}{\cot \theta + \cot \phi}

= tan ⁡ θ + tan ⁡ ϕ 1 tan ⁡ θ + 1 tan ⁡ ϕ \large= \dfrac{\tan \theta + \tan \phi}{\frac{1}{\tan \theta}+\frac{1}{\tan \phi}}

= tan ⁡ θ + tan ⁡ ϕ tan ⁡ ϕ + tan ⁡ θ tan ⁡ θ tan ⁡ ϕ \large= \dfrac{\tan \theta + \tan \phi}{\dfrac{\tan \phi + \tan \theta}{\tan \theta~\tan \phi}}

= tan ⁡ θ + tan ⁡ ϕ 1 × tan ⁡ θ tan ⁡ ϕ tan ⁡ θ + tan ⁡ ϕ \large=\dfrac{\tan \theta + \tan \phi}{1} \times \dfrac{\tan \theta~\tan \phi}{\tan \theta + \tan \phi}

= \large = tan ⁡ θ tan ⁡ ϕ \boxed{\large \tan \theta~\tan \phi}

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