Question

what is the value of tan 2 theta​

Answer 1

Answer:

see the attachment above you will understand

Answer 2

Answer:

The value of tan(2θ) is 2tanθ/(1-tan^2(θ)).

Step-by-step explanation:

We know that tan(x) = sin(x)/cos(x). Therefore,

$tan (2x) = \frac{sin(2x)}{cos(2x)}$

Using the double angle formula for sin and cos, we get

$tan(2x) = \frac{2sin(x)cos(x)}{cos^2(x)-sin^2(x)}$

Dividing the numerator and denominator by $cos^2(x)$

$tan(2x) = \frac{\frac{2sin(x)cos(x)}{cos^2(x)} }{\frac{cos^2(x)-sin^2(x)}{cos^2(x)} }$

$tan(2x) = \frac{\frac{2sin(x)cos(x)}{cos(x)cos(x)} }{\frac{cos^2(x)}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)} }$

$tan(2x)=\frac{\frac{2sin(x)}{cos(x)} }{1-\frac{sin^2(x)}{cos^2(x)} }$

$tan(2x) = \frac{2tan(x)}{1-tan^2(x)}$

Alternative method:

We can write tan(2x) = tan(x + x)

Using the formula $tan(x+y) = \frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$, we get

$tan(2x) = tan(x+x)$

$tan(2x) = \frac{tan(x)+tan(x)}{1-tan(x)tan(x)}$

$tan(2x) = \frac{2tan(x)}{1-tan^2(x)}$

Therefore, tan(2θ) = 2tanθ/(1-tan^2(θ))

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