Question

Prove that
1 - Cos 2A + sin 2A / 1 + cos 2A + sin 2A = tan A
​

Answer 1

Step-by-step explanation:

To Prove :-

$\rm  \dfrac{ 1 - cos2A +sin2A }{1 + cos2A + sin2A} = tanA$

Formulas used:-

→ cos2A = 1 - 2sin²A

→ cos2A = 2cos²A - 1

→ sin2A = 2sinA cosA

→ sinA/cosA = tanA

Proof:-

$\rm  LHS = \dfrac{ 1 - cos2A +sin2A }{1 + cos2A + sin2A}$

$= \rm\dfrac{ 1 -(1 - 2 {sin}^{2} A) +sin2A }{1 + (2 {cos}^{2} A - 1)+ sin2A}$

$= \rm\dfrac{ 1 -1 + 2 {sin}^{2} A+sin2A }{1 + 2 {cos}^{2} A - 1+ sin2A}$

$= \rm\dfrac{ 1 -1 + 2 {sin}^{2} A+sin2A }{1 - 1 + 2 {cos}^{2} A+ sin2A}$

$= \rm\dfrac{ 2 {sin}^{2} A+sin2A }{ 2 {cos}^{2} A+ sin2A}$

$= \rm\dfrac{ 2 {sin}^{2} A+2sinAcosA }{ 2 {cos}^{2} A+ 2sinAcosA}$

$= \rm\dfrac{ 2 {sin} A(sinA+cosA )}{ 2 {cos} A(sinA+cosA )}$

$= \rm\dfrac{ 2 {sin} A \cancel{(sinA+cosA )}}{ 2 {cos} A \cancel{(sinA+cosA )}}$

$= \rm\dfrac{ 2 {sin} A}{ 2 {cos} A}$

$= \rm tanA$

= RHS

LHS = RHS

Hence Proved