Question

If tan theta - tan(90 - theta) = 1 , then find the value of theta ?

Answer 1

Answer:

$\theta = tan^{-1}( \frac{1-\sqrt{5}}{2}), tan^{-1}(\frac{1+\sqrt{5}}{2})$

Step-by-step explanation:

It is given that

   $tan\theta - tan(90-\theta) = 1$

We know that $tab(90-\theta) = cot\theta$

⇒ $tan\theta-cot\theta = 1$

⇒ $tan\theta = 1 + cot\theta$

We know that $cot\theta = \frac{1}{tan\theta}$

⇒ $tan\theta = 1 + \frac{1}{tan\theta}$

⇒ $tan\theta = \frac{tan\theta + 1}{tan\theta}$

⇒ $tan^{2}\theta = tan\theta + 1$

Let $tan\theta = x$

⇒ $x^{2} = x + 1$

⇒ $x^{2} -x -1 = 0$

Applying quadratic formula

⇒ $x =$ $\frac{-b-\sqrt{b^{2}-4ac }}{2a} , \frac{-b+\sqrt{b^{2}-4ac }}{2a}$

$b^{2} - 4ac = (-1)^{2} -4(1)(-1) = 1 + 4 = 5$

⇒ $x = \frac{-(-1)-\sqrt{5}}{2}, \frac{-(-1)+\sqrt{5}}{2}$

⇒ $x = \frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}$

⇒ $tan \theta = \frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}$

⇒ $\theta = tan^{-1}( \frac{1-\sqrt{5}}{2}), tan^{-1}(\frac{1+\sqrt{5}}{2})$