Question

Coser At1 =(see A+tan A) ²
Corec A-1​

Answer 1

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:\dfrac{cosecA + 1}{cosecA - 1} = {(secA + tanA)}^{2}$

Identities Used :-

$\red{\rm :\longmapsto\:cosecx = \dfrac{1}{sinx}}$

$\red{\rm :\longmapsto\: {sin}^{2}x + {cos}^{2}x = 1}$

$\red{\rm :\longmapsto\:secx = \dfrac{1}{cosx}}$

$\red{\rm :\longmapsto\:tanx = \dfrac{sinx}{cosx}}$

$\large\underline{\sf{Solution-}}$

Consider, LHS

$\rm :\longmapsto\:\dfrac{cosecA + 1}{cosecA - 1}$

$\rm \: = \: \: \dfrac{\dfrac{1}{sinA} + 1 }{\dfrac{1}{sinA} - 1}$

$\rm \: = \: \: \dfrac{\dfrac{1 + sinA}{sinA} }{\dfrac{1 - sinA}{sinA}}$

$\rm \: = \: \: \dfrac{1 + sinA}{1 - sinA}$

$\rm \: = \: \: \dfrac{1 + sinA}{1 - sinA} \times \dfrac{1 + sinA}{1 + sinA}$

$\rm \: = \: \: \dfrac{ {(1 + sinA)}^{2} }{ {1}^{2} - {sin}^{2}A}$

$\rm \: = \: \: \dfrac{ {(1 + sinA)}^{2} }{{cos}^{2}A}$

$\rm \: = \: \: {\bigg(\dfrac{1 + sinA}{cosA} \bigg) }^{2}$

$\rm \: = \: \: {\bigg(\dfrac{1}{cosA} + \dfrac{sinA}{cosA} \bigg) }^{2}$

$\rm \: = \: \: {(secA + tanA)}^{2}$

Thus,

$\bf :\longmapsto\:\dfrac{cosecA + 1}{cosecA - 1} = {(secA + tanA)}^{2}$

Hence, Proved

Additional Information :-

$\red{\rm :\longmapsto\: {sec}^{2}x - {tan}^{2}x = 1}$

$\red{\rm :\longmapsto\: {cosec}^{2}x - {cot}^{2}x = 1}$

$\red{\rm :\longmapsto\:cotx = \dfrac{cosx}{sinx}}$

$\red{\rm :\longmapsto\:sin(90 \degree \: - x) = cosx}$

$\red{\rm :\longmapsto\:cos(90 \degree \: - x) = sinx}$

$\red{\rm :\longmapsto\:cosec(90 \degree \: - x) = secx}$

$\red{\rm :\longmapsto\:sec(90 \degree \: - x) = cosecx}$

$\red{\rm :\longmapsto\:tan(90 \degree \: - x) = cotx}$

$\red{\rm :\longmapsto\:cot(90 \degree \: - x) = tanx}$