Question

If x and y are complementary angles, then
(a) sin x = sin y
(b) tan x = tan y
(c) cos x = cos y
(d) sec x = cosec y​

Answer 1

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x and y are complementary => x+y= 90°

90°=> sin x = sin (90°- y) = cos y

90°=> sin x = sin (90°- y) = cos y⇒sin y = cos(90-x) = cos x

90°=> sin x = sin (90°- y) = cos y⇒sin y = cos(90-x) = cos xAlso, tan y = Cot (90 - y) = Cot x sec x = Cosec (90 - x) = cosec y

90°=> sin x = sin (90°- y) = cos y⇒sin y = cos(90-x) = cos xAlso, tan y = Cot (90 - y) = Cot x sec x = Cosec (90 - x) = cosec ysin y *tan x = cos x * sinx/cos x = sin x

90°=> sin x = sin (90°- y) = cos y⇒sin y = cos(90-x) = cos xAlso, tan y = Cot (90 - y) = Cot x sec x = Cosec (90 - x) = cosec ysin y *tan x = cos x * sinx/cos x = sin xtan y* cos x = cot y *cos x = cos y / sin y *sin y

90°=> sin x = sin (90°- y) = cos y⇒sin y = cos(90-x) = cos xAlso, tan y = Cot (90 - y) = Cot x sec x = Cosec (90 - x) = cosec ysin y *tan x = cos x * sinx/cos x = sin xtan y* cos x = cot y *cos x = cos y / sin y *sin y= cos y = sin x

90°=> sin x = sin (90°- y) = cos y⇒sin y = cos(90-x) = cos xAlso, tan y = Cot (90 - y) = Cot x sec x = Cosec (90 - x) = cosec ysin y *tan x = cos x * sinx/cos x = sin xtan y* cos x = cot y *cos x = cos y / sin y *sin y= cos y = sin xcosec y = 1/sin y = 1/cos x = sec x

hope it was helpful to you