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Question
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
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Solution
LHS =`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))`
=`((sinA - sin B )( sinA + sinB )+ ( cos A - cosB )( cosA - cosB))/((cos A+ cos B )( sin A+ sinB))`
=` (sin^2 A - sin^2 B + cos^2 A - cos^2 B)/( (cos A + cos B )( sinA + sinB))`
=` 0/((cos A + cos B )( sin A + sinB ))`
=0
=RHS
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L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
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= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
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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`
