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Question
If 3 cos A = 4 sin A, find the value of :
(i) cos A(ii) 3 - cot2 A + cosec2A.
Solution
Consider the diagram below :
3cos A = 4 sin A
cot A = `(4)/(3)`
i.e. `"base"/"perpendicular" = (4)/(3) ⇒ "AB"/"BC" = (4)/(3)`
Therefore if length of AB = 4x, length of BC = 3x
Since
AB2 + BC = AC2 ...[ Using Pythagoras Theorem]
(4x)2 + (3x)2 = AC2
AC2 = 25x2
∴ AC = 5x ...( hypotenuse)
(i) cos A = `"AB"/"AC" = (4)/(5)`
(ii) cosec A = `"AC"/"BC" = (5)/(3)`
Therefore
3–cot2 A + cosec2 A
= `3 – (4/3)^2+(5/3)^2`
= `(27 – 16 + 25)/(9)`
=`(36)/(9)`
= 4
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