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Show that cot(A + 15°) – tan(A – 15°) = AA4cos2A1+2sin2A

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प्रश्न

Show that cot(A + 15°) – tan(A – 15°) = `(4cos2"A")/(1 + 2 sin2"A")`

योग
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उत्तर

L.H.S = `(cos("A" + 15^circ))/(sin("A" + 15^circ)) - (sin("A" - 15^circ))/(cos("A" - 15^circ))`

= `(cos("A" + 15^circ)cos("A" - 15^circ) - sin("A" + 15^circ)sin("A" - 15^circ))/(sin("A" + 15^circ) cos("A" - 15^circ))`

= `(cos("A" + 15^circ + "A" - 15^circ))/(1/2[sin("A" + 15^circ + "A" - 15^circ) + sin("A" + 15^circ - "A" + 15^circ)]`

= `(2cos2"A")/(sin2"A" + sin30^circ)`

= `(2cos2"A")/(1/2 + sin2"A")`

= `(4cos2"A")/(1 + 2sin 2"A")`

= R.H.S

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Trigonometric Functions and Their Properties
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.6 [पृष्ठ १२२]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 3 Trigonometry
Exercise 3.6 | Q 14 | पृष्ठ १२२

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