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Question
If 3 `cot theta = 4 , "write the value of" ((2 cos theta - sin theta))/(( 4 cos theta - sin theta))`
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Solution
W e have ,
3 `cot theta = 4 `
⇒ ` cot theta = 4/3 `
Now,
`((2 cos theta + sin theta ))/((4 cos theta - sin theta))`
=` (((2 cos theta )/ sin theta + sin theta / sin theta))/(((4 cos theta) / sin theta - sin theta/ sin theta))` (๐ท๐๐ฃ๐๐๐๐๐ ๐๐ข๐๐๐๐๐ก๐๐ ๐๐๐ ๐๐๐๐๐๐๐๐๐ก๐๐ ๐๐ฆ sin ๐)
=`((2 cot theta +1))/((4 cot theta -1))`
=`((2xx4/3 +1))/((4xx4/3-1))`
=`((8/3+1/1))/((16/3-1/1))`
=`(((8+3)/3))/(((16-3)/3))`
=`((11/3))/((13/3))`
=`11/13`
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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`
