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Question
Prove that `(1 + sin B)/(cos B) + (cos B)/(1 + sin B) = 2 sec B`.
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Solution
L.H.S = `(1 + sin B)/(cos B) + (cos B)/(1 + sin B)`
= `((1 + sin B)^2 + cos^2B)/(cos B(1 + sin B))`
= `(1 + 2 sin B + sin^2B + cos^2B)/(cos B(1 + sin B))` ...[∵ (a + b)2 = a2 + 2ab + b2]
= `(1 + 2 sin B + 1)/(cos B(1 + sin B))` ...[∵ sin2B + cos2B = 1]
= `(2 + 2 sin B)/(cos B(1 + sin B))`
= `(2(1 + sin B))/(cos B(1 + sin B))`
= `2/(cos B)`
= 2 sec B
= R.H.S.
∴ `(1 + sin B)/(cos B) + (cos B)/(1 + sin B) = 2 sec B`
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
