1 + tansquare(45 - A)/1 - tansquare(45 - A) = cosec2A
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e 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.
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