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Question
If `cos theta = 2/3 , " write the value of" (4+4 tan^2 theta).`
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Solution
`4+4 tan^2 theta `
= `4(1+ tan ^2 theta)`
=`4 sec^2 theta `
=`4/ cos^2 theta`
=`4/(2/3)^2`
=`4/((4/9))`
=`(4xx9)/4`
=9
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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`
