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प्रश्न
Given: 4 cot A = 3
find :
(i) sin A
(ii) sec A
(iii) cosec2A - cot2A.
उत्तर
Consider the diagram below :
4 cot A = 3
cot A = `(3)/(4)`
i.e.`"base"/"perpendicular" = (3)/(4) ⇒ "AB"/"BC" =(3)/(4)`
Therefore if length of AB = 3x, length of BC = 4x
Since
AB2 + BC2 = AC2 ...[ Using Pythagoras Theorem ]
(3x)2 + (4x)2 = AC2
AC2 = 9x2 + 16x2 = 25x2
∴ AC = 5x ...( hypotenuse )
(i) sin A = `"perpendicular"/"hypotenuse " = (4x)/(5x) = (4)/(5)`
(ii) sec A = `"hypotenuse"/"base" = "AC"/"AB" = (5x)/(3x) = 5/3`
(iii) cosec A = `"hypotenuse"/"perpendicular" = "AC"/"BC" = (5x)/(4x) = (5)/(4)`
cot A = `(3)/(4)`
cosec2 A – cot2 A
=`(5/4)^2 – (3/4)^2`
= `( 25 - 9)/(16)`
= `(16)/(16)`
= 1
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