Given that $tan (\frac{1}{2} θ) = - 3$ $tan (1/2 theta) = -3$ . Show that $tan θ = \frac{3}{4}$ $tan theta = 3/4$ .?

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Answer 1 · 2018-04-12T17:25:36

We know the identity:

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

Let $x = \frac{1}{2} θ$ $x = 1/2theta$ , then $2 x = θ$ $2x = theta$ :

$tan (θ) = 2 \frac{tan (\frac{1}{2} θ)}{1 - {tan}^{2} (\frac{1}{2} θ)}$ $tan(theta) = 2tan(1/2theta)/(1-tan^2(1/2theta))$

Substitute $tan (\frac{1}{2} θ) = - 3$ $tan(1/2theta) = -3$ :

$tan (θ) = 2 \frac{- 3}{1 - {(- 3)}^{2}}$ $tan(theta) = 2(-3)/(1-(-3)^2)$

$tan (θ) = 2 \frac{- 3}{- 8}$ $tan(theta) = 2(-3)/(-8)$

$tan (θ) = \frac{3}{4}$ $tan(theta) = 3/4$ Q.E.D.

Given that tan(12θ)=−3tan (1/2 theta) = -3. Show that tanθ=34tan theta = 3/4.?