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Given : 5 cos A - 12 sin A = 0; evaluate: AAAAsinA+cosA2cosA–sinA - Mathematics

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प्रश्न

Given : 5 cos A - 12 sin A = 0; evaluate:

`(sin "A"+cos"A")/(2 cos"A"– sin"A")`

योग

उत्तर

5 cos A – 12 sin A = 0

5 cos A = 12 sin A

`sin "A"/cos "A" = (5)/(12)`

tan A = `(5)/(12)`

Now,

`(sin "A"+cos"A")/(2 cos"A"– sin"A") =  (sin"A"/cos"A" + cos"A"/cos"A")/(2 cos"A"/cos"A" – sin"A"/cos"A")`

= `(tan "A"+1)/(2– tan "A")` 

= `(5/12+1)/(2–5/12)`

= `(17/12)/(19/12)`

= `(17)/(19)`

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अध्याय 22: Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals] - Exercise 22 (B) [पृष्ठ २८६]

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सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 22 Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals]
Exercise 22 (B) | Q 14 | पृष्ठ २८६
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