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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

If ∆ABC is a right triangle and if ∠A = π2 then prove that cos B – cos C = BC-1+22cos B2 sin C2

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प्रश्न

If ∆ABC is a right triangle and if ∠A = `pi/2` then prove that cos B – cos C = `- 1 + 2sqrt(2) cos  "B"/2  sin  "C"/2`

बेरीज
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उत्तर

A + B + C = 180°

Given A = 90°

∴ B + C = 90°

⇒ `("B" + "C")/2` = 45°

`"B"/2 + "C"/2` = 45°

R.H.S = `- 1 + 2sqrt(2) cos  "B"/2 sin  "C"/2`

= `1 + sqrt(2) (2cos  "B"/2 sin  "C"/2)`

We know that 2 cosA sinB = sin(A + B) – sin(A – B)

= `- 1 + sqrt(2) (sin  (("B" + "C"))/2 - sin  (("B" - "C"))/2)`

= `- 1 + sqrt(2) (sin 45^circ - sin  (("B" - "C"))/2)`

= `- 1 + sqrt(2) (1/sqrt(2) - sin  (("B" - "C")^2)/2)`

= `- 1 + 1 - sqrt(2) sin  (("B" - "C"))/2`

= `- sqrt(2) sin  (("B" - "C"))/2`  .....(1)

L.H.S = cos B – cos C

= `2 sin  (("B" + "C"))/2 sin  (("C" - "B"))/2`

= `2 sin 45^circ sin  (("C" - "B"))/2`

= `2(1/sqrt(2)) sin  ((-("B" - "C"))/2)`

= `- sqrt(2) sin  (("B" - "C"))/2` .....(2)

From (1) and (2)

⇒ L.H.S = R.H.S

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Trigonometric Functions and Their Properties
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometry - Exercise 3.7 [पृष्ठ १२४]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 3 Trigonometry
Exercise 3.7 | Q 5. (iii) | पृष्ठ १२४

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