Question

using the formula,tan 2A = 2tanA/1-tan^2A,find the value of tan
60°, it being given that tan 30° = 1/√3

pls help me to solve
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Answer 1

Step-by-step solution:

FORMULA :    $tan2A=\frac{2tanA}{1-tan^{2}A}$

Now, put   $A=30^{0}$

It becomes:

=>  $tan[2(30^{0})]=\frac{2tan(30^{0})}{1-tan^{2}(30^{0})}$

And it is given that,   $tan(30^{0})=\frac{1}{\sqrt{3} }$

=>  $tan(60^{0})=\frac{2(\frac{1}{\sqrt{3} }) }{1-(\frac{1}{\sqrt{3} })^{2}}$

                    $=\frac{2\sqrt{3} }{3-1}$

=>  $tan(60^{0})= \sqrt{3}$

Hence, Proved.

Answer 2

Answer:

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☜☆Explain☆☞

Why the electron gain enthalpy of Chlorine is greater than that of Flourine????

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