Question

using Pythagoras theorem prove that 1+cot²theta = cosec² theta​

Answer 1

To prove:

1+cot²theta = cosec² theta using Pythagora's Theorem.

Proof:

Let us consider a ∆ABC such that AB = a and BC = b and $\angle ABC$ = 90°.

Now, as per Pythagora's Theorem:

$\therefore \: {(AB)}^{2} + {(BC)}^{2} = {(AC)}^{2}$

$= > \: {a}^{2} + {b}^{2} = {(AC)}^{2}$

$= > \: AC = \sqrt{ {a}^{2} + {b}^{2} }$

Now, LHS:

$\therefore \: 1 + { \cot}^{2} ( \theta)$

$= \: 1 + { (\dfrac{base}{altitude}) }^{2}$

$= \: 1 + { (\dfrac{b}{a}) }^{2}$

$= \: 1 + \dfrac{ {b}^{2} }{ {a}^{2} }$

$= \: \dfrac{ {a}^{2} + {b}^{2} }{ {a}^{2} }$

Now , RHS:

$\therefore \: { \csc }^{2} ( \theta)$

$= \: { (\dfrac{hypotenuse}{altitude}) }^{2}$

$= \: { \bigg(\dfrac{ \sqrt{ {a}^{2} + {b}^{2} } }{a} \bigg) }^{2}$

$= \: \dfrac{ {(\sqrt{ {a}^{2} + {b}^{2} })}^{2} }{ {a}^{2} }$

$= \: \dfrac{ {a}^{2} + {b}^{2} }{ {a}^{2} }$

So, LHS = RHS , hence proved.