Question

Prove: tan(62)=2tan(34)+tan(28)

Answer 1

Answer:

$tan62 = 2tan34+tan28$

Step-by-step explanation:

$tan62= tan(34+28)\\=\frac{tan34+tan28}{1-tan34tan28}$

$Since, \\\boxed {tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}}$

$\implies tan62(1-tan34tan28)=tan34+tan28$

$\implies tan62-tan28tan34tan62\\=tan34+tan28$

$\implies tan62-tan28tan34tan(90-28)\\=tan34+tan28$

$\implies tan62-tan28tan34cot28\\=tan34+tan28$

$\implies tan62-tan28cot28tan34\\=tan34+tan28$

$\implies tan62-1\times tan34\\=tan34+tan28$

/* Since, tanAcotA = 1*/

$\implies tan62 = tan34+tan28+tan34$

$\implies tan62 = 2tan34+tan28$

Therefore,

$tan62 = 2tan34+tan28$

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Answer 2

Answer:

tan 62° = 2 tan 34° + tan 28°

step by step explanation:

=> tan 62° = tan (34° + 28°)

=> tan 62° = (tan 34° + tan 28°) / (1 - tan 34° × tan 28°) [since, tan (A+B) = (tanA + tanB) / (1 - tanA × tanB)]

=> tan 62° × (1 - tan34° × tan28°) = tan 34° + tan 28°

=> tan 62° - tan 34° × tan 28° × tan 62° = tan 34° + tan 28°

=> tan 62° - tan 34° × tan 28° × tan (90° - 28°) = tan 34° + tan 28°

=> tan 62° - tan 34° × tan 28° × cot 28° = tan 34° + tan 28°

=> tan 62° - tan 34° × 1 = tan 34° + tan 28° [ since, tanA × cotA = 1]

=> tan 62° = tan 34° + tan 34° + tan 28°

=> tan 62° = 2 tan 34° + tan 28°

Hence, it is prove.

Hope it will help you