Question

what is the value of tab 15​

Answer 1

-0.8559934009. tan15

Answer 2

here is value of tan 15

Clearly, tan15°=tan(45°-30°)

We know that,

tan(A-B)=(tanA-tanB)/(1+tanA.tanB)

So, tan15°=tan(45°-30°)

=(tan 45°-tan30°)/(1+tan45°.tan30°)

=[{1-(1/√3)}/{1+(1)(1/√3)}]

=(√3–1)/(√3+1)

Rationalising the denominator, we have,

Tan15°= {(√3–1)×(√3–1)}/{(√3+1)×(√3–1)}

=(3+1–2√3)/(3–1)

=(4–2√3)/2

=2-√3.

Aliter

Let θ=15°

Then, tanθ=tan15°

So, tan2θ=tan2(15°)=tan30°

We know that,

Tan2θ=2tan theta/(1-tan^2 theta)

=>Tan30°=2tan15°/(1-tan^2 15°)

=>1/√3 =2tan15°/(1-tan^2 15°)

=>1-tan^2 15=2√3 tan 15°

=>tan^2 15° + 2√3 tan15° - 1 = 0

Now, by Quadratic Formula, we have

=> Tan15°=[-2√3 ± √{(-2√3)^2 - 4(1)(-1)}]/2(1)

=(-2√3+√16)/2

=(4–2√3)/2

=2-√3