Question

4) If coso=2x/1+x^2,then tano=​

Answer 1

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Given :

CosØ = 2x/1 + x²

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To Find :

• TanØ

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Solution :

As we know that :

$\large{\boxed{\sf{\cos \theta \: = \: \dfrac{Base}{Hypotenuse}}}}$

So,

• Base = 2x
• Hypotenuse = 1 + x²

Now use Pythagoras theorem

${\boxed{\sf{Hypotenuse^2 \: = \: Base^2 \: + \: Perpendicular^2}}} \\ \\ \implies {\sf{(1 \: + \: x^2)^2 \: = \: (2x)^2 \: + \: Perpendicular^2}} \\ \\ \implies {\sf{1 \: + \: x^4 \: + \: 2x^2 \: = \: 4x^2 \: + \: Perpendicular^2}} \\ \\ \implies {\sf{Perpendicular^2 \: = \: 1 \: + \: x^4 \: + \: 2x^2 \: - \: 4x^2}} \\ \\ \implies {\sf{Perpendicular^2 \: = \: 1 \: + \: x^4 \: - \: 2x^2}} \\ \\ \implies {\sf{Perpendicular \: = \: \sqrt{1 \: + \: x^4 \: - \: 2x^2}}}$

Now, we know that :

• TanØ = Perpendicular/Base

⇒TanØ = $\sf{\dfrac{\sqrt{(1 \: + \: x^4 \: - \: 2x^2)}}{2x}}$