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प्रश्न
In the given figure, AC = 13cm, BC = 12 cm and ∠B = 90°. Without using tables, find the values of: `("cos A" - "sin A")/("cos A" + "sin A")`
उत्तर
ΔABC is a right-angled triangle.
∴ AC2 = AB2 + BC2
⇒ AB2
= AC2 - BC2
= 132 - 122
= 169 - 144
= 25
⇒ AB = 5cm
sin A = `"BC"/"AC" = (12)/(13)`
cos A = `"AB"/"AC" = (5)/(13)`
`("cos A" - "sin A")/("cos A" + "sin A")`
= `(5/13 - 12/13)/(5/13 + 12/13)`
= `(-(7)/(13))/((17)/(13)`
= `-(7)/(13) xx (13)/(17)`
= `-(7)/(17)`.
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