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प्रश्न
If 3 cot A = 2, then find the value of `(4sin"A" - 3cos"A")/(2sin"A" + 3cos"A")`
उत्तर
3 cot A = 2
⇒ cot A = `2/3`
AC2 = AB2 + BC2
= 32 + 22
= 9 + 4
AC = `sqrt(13)`
cos A = `"AB"/"AC" = 3/sqrt(13)`
sin A = `"BC"/"AC" = 2/sqrt(13)`
The value of `(4sin"A" - 3cos"A")/(2sin"A" + 3cos"A")`
= `4(2/sqrt(13)) - 3(3/sqrt(13)) ÷ 2(2/sqrt(13)) + 3(3/sqrt(13))`
= `(8/sqrt(13)) - (9/sqrt(13)) ÷ (4/sqrt(13)) + (9/sqrt(13))`
= `((8 - 9)/sqrt(13)) ÷((4 + 9)/sqrt(13))`
= `((-1)/sqrt(13)) ÷ (3/sqrt(13))`
= `((-1)/sqrt(13)) xx (sqrt(13)/13)`
The value is `((-1)/13)`
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