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प्रश्न
Given: sec A = `( 29 )/(21), "evaluate : sin A" - 1/tan "A"`
उत्तर
Consider the diagram below :
sec A = `(29)/(21)`
i.e. `"hypotenuse"/"base" = (29)/(21) ⇒ "AC"/"AB" = (29)/(21)`
Therefore if length of AB = 21x , length of AC = 29x
Since
AB2 + BC2 = AC2 ...[ Using Pythagoras Theorem ]
(21x)2 + BC2 = ( 29x)2
BC2 = 841x2 – 441x2 =400x2
BC = 20x ...(perpendicular )
Now
sin A = `"perpendicular"/"hypotenuse" = (20x)/(29x) = 20/29`
tan A = `"perpendicular"/"base" = (20x)/(21x) = 20/21`
Therefore
sin A – `1/ tan "A"`
= `(20)/(29) – 1/(20/21)`
= `(20)/(29) – 21/20`
= – `(209)/(580)`
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