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Question
Show that none of the following is an identity:
`tan^2 theta + sin theta = cos^2 theta`
Solution
`tan^2 theta + sin theta = cos^2 theta`
LHS = `tan^2 theta + sin theta `
=`(sin^2 theta)/(cos^2 theta) + sin theta`
=` (1- cos^2 theta )/( cos^2 theta) + sin theta`
=` sec^2 theta -1 + sin theta `
Since LHS ≠ RHS, this is not an identity.
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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`
