Prove that tan(45°+ A) - tan(45°- A)/ tan(45°+ A) + tan(45°- A) = 1/sin2A

To prove the above equation, we will use trigonometric identities and value of tan45°.

Answer: Hence Proved that tan(45°+A) - tan(45°- A) / tan(45°+A) + tan(45°-A) = 1/sin2A

Let us see the solution.

Explanation:

As we know that tan (A + B) = tan A + tan B/ 1 - tan A tan B and tan (A - B) = tan A - tan B = 1 + tan A tan B

Using this identity, we will solve LHS.

Substitute A = 45° and B = A in this formula.

LHS = tan (45°+ A) - tan(45°- A) / tan(45°- A) + tan(45° + A)                                   

tan (45°+ A) - tan(45°- A) = (1+ tanA)² / (1 – tanA)² + (1 - tanA)² / ( 1 – tanA)² 

= ((1+ tanA)² + (1 - tanA)²) / (1 – tanA)²

= 2 (1+ tanA)² / (1 – tanA)²

= 2 sec 2a

Similarly, tan (45°- A) + tan (45° + A)  = 2 tan 2a                                                                                                        

So, tan (45°+ A) - tan (45°- A) / tan (45°- A) + tan (45° + A) = 2 sec 2a / 2 tan 2a = 1/sin 2a

Thus, tan (45°+A) - tan(45°- A) / tan (45°+A) + tan (45°- A) = 1/sin 2a