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Questions
`1 + (tan^2 θ)/((1 + sec θ)) = sec θ`
Prove the following:
`1 + (tan^2 θ)/(1 + sec θ) = sec θ`
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Solution
LHS = `1 + (tan^2 θ)/((1 + sec θ))`
=` 1 + ((sec^2 θ - 1))/((sec theta + 1))`
=`1 + ((sec theta + 1)(sec theta - 1))/((sec theta + 1))`
=`1 + (sec theta - 1)`
= sec θ
LHS = RHS
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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`
