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Question
If \[\cos A = \frac{7}{25}\] find the value of tan A + cot A.
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Solution
Given: `cos A=7/25`
We know that,
`sin^2 A+cos^2 A=1`
⇒` sin^2 A+(7/25)^2=1`
⇒` sin^2 A+49/625=1`
⇒` sin^2 A1-49/625`
⇒ `sin^2A=576/625`
⇒ `sin A=24/25`
Therefore,
`tan A+cot A= sin A/cos A+cos A/sin A`
=` (24/25)/(7/25)+1=(7/25)/(24/25)`
= `24/7+7/24`
=`((24)^2+(7)^2)/168`
=`(576+49)/168`
=`625/168`
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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`
