Question

1-tan^2(45°-A)/1+tan^2(45°-A)=sin2A

Answer 1

Hey......!!! here is ur answer........☺️☺️☺️

We have to prove that.......

1–tan²(45°–A)/1+tan(45°–A) = sin2A

On taking L.H.S......

1–tan²(45°–A)/1+tan(45°–A)...........(1)

As we know that.....sec²A = 1+tan²A then from equation (1)

=>1–tan²(45°–A)/sec²(45°–A)

=>1/sec² (45°–A) – tan²(45–A)/sec²(45–A)

=>cos²(45–A) – sin²(45–A)

{ Because cosA = 1/secA and tanA/secA = sinA}

=>cos2(45°–A)

{ Because cos²A – sin²A = cos2A }

=>cos(90°–2A)

=>sin2A { Because cos(90–A) = sinA }

= R.H.S.

I hope it will help you.........✌️✌️✌️

Answer 2

refer to above attachment hope this helps you....

have a nice day.....