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Question
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
Solution
LHS = `(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))`
=`tan^2 theta/sec^2 theta + cot^2 theta/ cosec ^2 theta (∵ sec^2 theta - tan^2 theta = 1 and cosec^2 theta - cot^2 theta=1)`
=`(sin^2theta/cos^2 theta)/(1/cos^2theta) + (cos^2theta/sin^2 theta)/(1/sin^2 theta)`
=` sin^2 theta + cos^2 theta`
=1
= RHS
Hence, 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`
