Question

What is the exact value of tan(67.5°)?

1. square root 3
2. square root 3/3
3. square root 2-square root2 devided by 2 plus square root 2
4. square root 2+ square root2 devided by 2 minus square root 2​

Answer 1

$\underline{\textbf{To find:}}$

$\mathsf{The\;value\;of\;tan(67.5^\circ)}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Formula used:}}$

$\mathsf{cos\,2A=\dfrac{1-tan^2A}{1+tan^2A}}$

$\mathsf{Consider,\;cos\,135^\circ}$

$\mathsf{=cos(90^\circ+45^\circ)}$

$\mathsf{=-sin\,45^\circ}$

$\mathsf{=-\dfrac{1}{\sqrt{2}}}$

$\mathsf{Take,\;\theta=67.5^\circ}$

$\implies\mathsf{2\theta=135^\circ}$

$\implies\mathsf{cos\,2\theta=cos\,135^\circ}$

$\textsf{Using the above formula,}$

$\mathsf{\dfrac{1-tan^2\theta}{1+tan^2\theta}=-\dfrac{1}{\sqrt{2}}}$

$\mathsf{\sqrt{2}-\sqrt{2}\,tan^2\theta=-1-tan^2\theta}$

$\mathsf{\sqrt{2}+1=\sqrt{2}\,tan^2\theta-tan^2\theta}$

$\mathsf{\sqrt{2}+1=(\sqrt{2}-1)\,tan^2\theta}$

$\mathsf{tan^2\theta=\dfrac{\sqrt{2}+1}{\sqrt{2}-1}}$

$\mathsf{Multiplying\;both\;numerator\;and\;denominator\;by\;\sqrt{2}}$

$\mathsf{tan^2\theta=\dfrac{2+\sqrt{2}}{2-\sqrt2}}$

$\textsf{Taking square root on bothsides, we get}$

$\mathsf{tan\theta=\pm\sqrt{\dfrac{2+\sqrt{2}}{2-\sqrt2}}}$

$\implies\mathsf{tan\,67.5^\circ=\pm\sqrt{\dfrac{2+\sqrt{2}}{2-\sqrt2}}}$

$\mathsf{But\;67.5^\circ\;is\;acute,\;tan\,67.5^\circ\;is\;positive}$

$\implies\boxed{\mathsf{tan\,67.5^\circ=\sqrt{\dfrac{2+\sqrt{2}}{2-\sqrt2}}}}$

Answer 2

Answer:

The exact value is $tan(67.5)=\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2} } }$.

Step-by-step explanation:

Given trigonometric angle is $tan 67.5$°.

We need to find the value of the tan at that angle.

To find this value, we use the value of tan 135°.

We can see that,

$135=2\times 67.5$

Taking tan on both sides,

$tan(135)=tan(2\times 67.5)$

We know the value of tan 45° = 1

We can write,

$tan(135)=tan(90+45)$

$=-cot(45)$

Since $tan(90+x)=-cotx$.

⇒ $tan(135)=-1$

We know the formula,

$tan2A=\frac{2tanA}{1-tan^2A}$.

$tan(2\times67.5)=\frac{2tan(67.5)}{1-tan^267.5}$

$-1=\frac{2tan67.5}{1-tan^267.5}$

Cross multiply the above equation, we get

$-1+tan^267.5=2tan67.5$

$tan^267.5-2tan67.5-1=0$

$tan67.5=\frac{-(-2)+\sqrt{(-2)^2-4(-1)}}{2(1)}, \frac{-(-2)-\sqrt{(-2)^2-4(-1)}}{2(1)}$

$tan67.5=\frac{2+2\sqrt{2}}{2}, \frac{2-2\sqrt{2}}{2}$

$tan67.5=1+\sqrt{2}, 1-\sqrt{2}$

Since $tan67.5$ is positive, because it is in 1st Quadrant.

So the value is

$tan67.5=1+\sqrt{2}$.

Now in option 4) the answer is $\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}}$.

Rationalising the numerator and denominator, we get

$\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}\times \frac{2+\sqrt{2}}{2+\sqrt{2}}}$

$=\sqrt{\frac{(2+\sqrt{2})^2}{4-2} }$

$=\frac{2+\sqrt{2}}{\sqrt{2}}$

$=\sqrt{2}+1$

$=tan67.5$°

So the correct answer is 4.

$tan(67.5)=\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2} } }$.