How do you find the value of $tan 20 (θ)$ $tan 20(theta)$ using the double angle identity?

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Answer 1 · 2017-10-11T11:24:41

Have a look.

For solving this, one should be aware of the formula $\to$ $rarr$

$tan 2 θ = \frac{2 tan θ}{1 - {tan}^{2} θ}$ $tan2theta=(2tantheta)/(1-tan^2theta)$

So, $tan (20 θ)$ $tan(20theta)$

$= tan (2 \times 10 θ)$ $=tan(2xx10theta)$

$= \frac{2 tan (10 θ)}{1 - {tan}^{2} (10 θ)}$ $=(2tan(10theta))/(1-tan^(2)(10theta)$

$= \frac{2 tan (2 \times 5 θ)}{1 - {(tan (2 \times 5 θ))}^{2}}$ $=(2tan(2xx5theta))/(1-(tan(2xx5theta))^2)$

In this way continue till you get an expression having only the term $5 θ$ $5theta$ .

Hope it Helps:)

How do you find the value of tan20(θ)tan 20(theta) using the double angle identity?