Question

Tan 32 deg + tan 88 deg / 1 - tan32 tan88 = -√3

Answer 1

Explanation:

`\displaystyle\\(tan32^0+tan88^0)/(1-tan32^0tan88^0) \ \ \ \ (1)\\\\Let\ \alpha =32^0\ and\ \beta =88^0\\\\Hence,\ expression\ (1)\ will \ look\ like\ this:\\\\(tan\alpha +tan\beta )/(1-tan\alpha tan\beta )\\`

The trigonometric formula for adding angles for a tangent is as follows:

`\displaystyle\\\\\boxed {tan(\alpha +\beta )=(tan\alpha +tan\beta )/(1-tan\alpha tan\beta )}`

Thus,

`\displaystyle\\(tan32^0 +tan88^0 )/(1-tan32^0 tan88^0 )=tan(32^0+88^0)\\\\\\(tan32^0 +tan88^0 )/(1-tan32^0 tan88^0 )=tan120^0`

Calculate the value of tan120°:

`tan120^0=tan(90^0+30^0)\\\\tan120^0=-cot30&^0\\\\tan120^0=-√(3)`

So,

`\displaystyle\\(tan32^0 +tan88^0 )/(1-tan32^0 tan88^0 )\equiv-√(3)`