Question

tan a + b is equal to tan A + tan b by 1 minus tan a tan B so that tan 75 is equal to 2 + root 3​

Answer 1

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We are given,

$\large{\boxed{\sf{\tan(A \: + \: B) \: = \: \dfrac{\tan A \: + \: \tan B}{1\: - \: \tan A \tan B}}}}$

We have to show that, tan75 = 2 + √3

So, Take A = 45° and B = 30°

$\implies {\sf{\tan (45 \: + \: 30) \: = \: \dfrac{\tan 45 \: + \: \tan 30}{1 \: - \: \tan 45 \tan 30}}} \\ \\ \implies {\sf{\tan(45 \: + \: 30) \: = \: \dfrac{1 \: + \: \dfrac{1}{\sqrt{3}}}{1 \: - \: 1 \times \dfrac{1}{\sqrt{3}}}}} \\ \\ \implies {\sf{\tan (45 \: + \: 30) \: = \: \dfrac{ \dfrac{\sqrt{3} \: + \: 1}{\sqrt{3}}}{\dfrac{\sqrt{3} \: - \: 1}{\sqrt{3}}}}} \\ \\ \implies {\sf{\tan (45 \: + \: 30) \: = \: \dfrac{\sqrt{3} \: + \: 1}{\sqrt{3} \: - \: 1}}} \\ \\ \implies {\sf{\tan(75) \: = \: \dfrac{\sqrt{3} \: + \: 1}{\sqrt{3} \: - \: 1} \: \times \: \dfrac{\sqrt{3} \: + \: 1}{\sqrt{3} \: + \: 1}}} \\ \\ \implies {\sf{\tan (75) \: = \: \dfrac{4 \: + \: 2 \sqrt{3}}{2}}} \\ \\ \implies {\sf{\tan (75) \: = \: \dfrac{2(2 \: + \: \sqrt{3})}{2}}} \\ \\ \implies {\sf{\tan (75) \: = \: 2 \: + \: \sqrt{3}}}$

Answer 2

Explanation:

Please find the ATTACHMENT.