Question

√(1-cos2A/1+cos2A)=tanA​

Answer 1

Answer:

Step-by-step explanation:

√(1-cos2A/1+cos2A)

//Multiply numerator and denominator with 1 - cos2A

=> √(1-cos2A/1+cos2A) * (1 - cos2A/1 - cos2A)

=> √ (1-cos2A)²/ 1 - cos²2A

=> 1 - cos2A/Sin2A

//Cos2A = cos²A - sin²A and Sin2A = 2SinACosA

=> sin²A + cos²A - (cos²A - sin²A) / 2sinAcosA

=> 2 sin²A/2sinAcosA

=> TanA

= R.H.S

Hence proved.

Answer 2

Answer:

Step-by-step explanation:

We know that,

cos2A = 2cos²A - 1 = 1 - 2sin²A

Now, $\sqrt{\frac{1-cos2A}{1+cos2A} } = \sqrt{\frac{1-1+2sin^{2}A }{1+2cos^{2}A-1 } }$

                         $=\sqrt{\frac{2sin^{2}A }{2cos^{2}A } }$

                         $=\frac{sinA}{cosA}$

                         $= tanA$

Thus, $\sqrt{\frac{1-cos2A}{1+cos2A} } = tanA$