Question

prove that tan 70°=2tan 50°+tan 20°​

Answer 1

Step-by-step explanation:

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$prove \: \: that \implies \\ \\ \tan(70) = 2 \tan(50) + \tan(20)$

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$\huge\red{Formula:-}$

$\boxed {\leadsto \tan( \alpha + \beta ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{1 - \tan( \alpha ) \tan( \beta ) } } \\ \\ \boxed{ \leadsto \tan( \theta) \times \cot( \theta) = 1}$

$We \: \: know... \\ \\ \tan(70) = \tan(50 + 20) \\ \\ \implies \tan(70) = \frac{ \tan(50) + \tan(20) }{1 - \tan(50) \tan(20) } \\ \\ \implies \tan(70) \times (1 - \tan(50) \tan(20) \tan(70 ) ) = \tan(50) + \tan(20) \\ \\ We \: \: know, \\ \\ \large \boxed{ \tan(20) = \cot(90 - 20) = \cot(70) } \\ \\ \implies \tan(70) - \tan(50) = \tan(50) + \tan(20) \\ \\ \huge \boxed{ \implies \tan(70) = 2 \tan(50) + \tan(20) }$

More basic identities..

$\\ 1) { \sin}^{2}(\theta) + {\cos}^{2}(\theta) = 1 \\ \\ 2) {\sec}^{2}(\theta) - {\tan}^{2} (\theta) = 1 \\ \\ 3) {\csc}^{2} (\theta) - {\cot}^{2} (\theta) = 1 \\ \\ 4) \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \\ \\ 5) \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} \\ \\ 6) \cot(\theta) = \frac{1}{\tan(\theta)} \\ \\ 7) \sin(\theta) = \frac{1}{\csc(\theta)} \\ \\ 8) \cos(\theta) = \frac{1}{\sec(\theta)}$

Answer 2

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