Seems like there is a correction in the first question (RHS is tanx.tany)
(i) For convenience: let tanx = a ; tany = b
Thus, tanx + tany = a + b
Moreover, cotx = 1/tanx = 1/a ; coty = 1/b
Thus,
cotx + coty = 1/a + 1/b = (a + b)/ab
Hence,
=> (tanx + tany)/(cotx + coty)
=> (a + b) / { (a + b)/ab }
=> ab(a + b)/(a + b)
=> ab => tanx.tany , proved.
(ii) For convenience: let sinx = a ; cosx = b
As we know, sin²x + cos²x = 1 => a²+b²=1
=> (a³ + b³)/(a + b)
=> (a + b)(a² + b² - ab) / (a + b)
=> (a² + b² - ab)
=> 1 - ab => 1 - sinx.cosx , proved
(iii): let x/2 = A
=> tan(x/2) + cosx.tan(x/2)
=> tanA + cos2A.tanA
=> tanA [1 + cos2A]
=> tanA (2cos²A) {1+cos2A = 2cos²A}
=> (sinA/cosA) (2cos²A)
=> sinA (2cosA)
=> 2sinAcosA
=> sin2A
=> sin2(x/2)
=> sinx proved
Letting x/2 = A is not mandatory. I did it to decease words*(in a line).
<u>Indentities used</u>:
• sin²A + cos²A = 1
• (a³ + b³) = (a + b)(a² + b² - 1)
• 1 + cosA = 2 cos²(A/2)
• tanA = sinA/cosA.
• 2sinAcosA = sin2A
