Question

cos2A\1-sin2A = 1 + tanA\1 - tanA​

Answer 1

To prove --->

Cos2A / 1 - Sin2A = 1+ tanA / 1 - tanA

Proof--->

We have some formulee

Cos 2A = Cos²A - Sin²A

Sin²A + Cos²A = 1

Sin2A = 2 SinA CosA

Now we find value of Cos2A

Cos2A = Cos²A - Sin²A

= (CosA )² - ( SinA )²

We have an identity

a² - b² = ( a + b ) ( a - b )

Applying this here

Cos2A = ( CosA + SinA ) ( CosA - SinA )

Now we simplify denominator

1 - Sin2A = Sin²A + Cos²A - 2 SinA CosA

= ( CosA - SinA )²

LHS

Cos2A

=--------------------------

1 - Sin 2A

(CosA + SinA ) ( CosA - SinA )

= ------------------------------------------------

( CosA - SinA )²

( CosA - SinA ) cancel out from numerator and denominator

( CosA + SinA )

= ----------------------------

( CosA - SinA )

Dividing by CosA in numerator and denominator

CosA /CosA + SinA / CosA

= ----------------------------------------------

CosA / CosA - SinA / CosA

using SinA / CosA = tanA and

CosA / CosA = 1

1 + tanA

= ----------------- = RHS

1 - tanA

Answer 2

$\frac{cos2a}{1 - sin2a} = \frac{1 + tana}{1 - tana}$

Here, use formula

• cos2a= (1-tan^2a)/(1+tan^2a)

and

• sin2a = 2tana/ 1+tan^2a

Put these two in LHS , we get

$\rightarrow \:\huge{ \frac{ \frac{1 - {tan}^{2} a}{1 + {tan}^{2} a} }{1 - \frac{2tana}{1 + {tan}^{2}a } }} \\ \\ \rightarrow \: \huge{ \frac{ \frac{1 - {tan}^{2}a }{1 + {tan}^{2} a} }{ \frac{1 + {tan}^{2} a - 2 tana }{1 + {tan}^{2} a} }}$

Here , 1+tan^2 a gets cancel,

and In denominator (1+tan^2a-2tana) = ( 1- tana ) ^2

and in numerator we write 1-tan^2a = ( 1+tana) ( 1-tana)

$\rightarrow \huge{ \: \frac{(1 + tana)(1 - tana)}{( {1 - tana})^{2} }}$

Here , ( 1- tana ) in both num and deno gets cancel , we get

$\huge \: \rightarrow \: \frac{1 + tana}{1 - tana}$

LHS = RHS

$\huge\boxed{ \purple{ \mathfrak{proved}}}$