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Question
`(tan A + tanB )/(cot A + cot B) = tan A tan B`
Solution
LHS = `(tan A + tanB )/(cot A + cot B) `
=`(tan A + tan B)/(1/ tan A + 1/ tanB)`
=` (tan A + tan B)/( (tan A+tan B)/ (tan A tan B)`
=`(tan A tan B ( tan A + tan B))/((tan A + tan B ))`
= ๐ก๐๐๐ด ๐ก๐๐๐ต
= 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`
