Question

If tano + coto = 2, then show that tan20 + cot20 = 2​

Answer 1

Given

$\sf \to \: tan \theta + cot \theta = 2$

To Prove

$\sf \to \: tan {}^{2} \theta + cot {}^{2} \theta = 2$

Now Take

$\sf \to \: tan \theta + cot \theta = 2$

Squaring on both side

$\sf \to \: (tan \theta + cot \theta) {}^{2} = (2) {}^{2}$

By using this identities

$\sf \to \: (a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab$

We get

$\sf \to \: tan {}^{2} \theta + cot {}^{2} \theta + 2 \times tan \theta \times cot \theta = 4$

We Know That

$\sf \to \: tan \theta = \dfrac{sin \theta}{cos \theta} \\ \\ \sf \to \: cot \theta = \frac{cos \theta}{ sin \theta}$

Now

$\sf \to \: tan {}^{2} \theta + cot {}^{2} \theta + 2 \times \dfrac{sin \theta}{cos \theta} \times \dfrac{cos \theta}{sin \theta} = 4$

$\sf \to \: tan {}^{2} \theta + cot {}^{2} \theta + 2 = 4$

$\sf \to \: tan {}^{2} \theta + cot {}^{2} \theta= 4 - 2$

$\sf \to \: tan {}^{2} \theta + cot {}^{2} \theta= 2$

Hence Proved