Maharashtra State BoardHSC Science (General) 11th
Advertisement Remove all ads

Prove the following: cosec 48° + cosec 96° + cosec 192° + cosec 384° = 0 - Mathematics and Statistics

Advertisement Remove all ads
Advertisement Remove all ads
Sum

Prove the following:

cosec 48° + cosec 96° + cosec 192° + cosec 384° = 0

Advertisement Remove all ads

Solution

cosec 48° = cosec (90° – 42°) = sec 42°

= `1/cos42^circ`  ...(1)

cosec 192° = cosec (270° – 78°) = – sec 78°

= `-1/cos78^circ` ...(2)

cosec 384 ° = cosec (360° + 24°) = cosec 24° ...(3)

∴ cosec 48° + cosec 192°

= `1/(cos42^circ) - 1/cos78^circ` ...[By (1) and (2)]

= `(cos78^circ -  cos42^circ)/(cos78^circ*  cos42^circ)`

= `(-2 sin ((78^circ + 42^circ)/2)*sin((78^circ - 42^circ)/2))/(cos(60^circ + 18^circ)*cos(60^circ - 18^circ)`

= `(-2  sin 60^circ*  sin18^circ)/(cos^2 60^circ - sin^2 18^circ)` ...[∵ cos(A + B) · cos(A – B) = cos2A – sin2B]

= `(-2 xx sqrt(3)/2 xx (sqrt(5) - 1)/4)/((1/2)^2 - ((sqrt(5) - 1)/4)^2)   ...[because sin18^circ = (sqrt(5) - 1)/4]`

= `(-sqrt(3)/4 (sqrt(5) - 1))/(1/4 - ((5 + 1 - 2sqrt(5))/16)`

= `(-sqrt(3)(sqrt(5) - 1))/4 xx 16/(4 - 6 + 2sqrt(5)`

=  `(-sqrt(3)(sqrt(5) - 1))/4 xx 16/(2(sqrt(5) - 1))`

∴ cosec 48° + cosec 192° = `-2sqrt(3)`  ...(4)

Also, cosec 96° + cosec 384°

= cosec 96° + cosec 24°  ...[By (3)]

= `1/(sin96^circ) + 1/sin24^circ`

= `(sin24^circ +  sin96^circ)/(sin96^circ*  sin24^circ)`

= `(2sin((96^circ + 24^circ)/2)*cos((96^circ - 24^circ)/2))/(sin(60^circ + 36^circ)*sin(60^circ - 36^circ)`

= `(2  sin60^circ*  cos36^circ)/(sin^2 60^circ - sin^2 36^circ)`  ...[∵ sin(A + B) · sin(A – B) = sin2A – sin2B]

= `(2 xx sqrt(3)/2 xx (sqrt(5) + 1)/4)/((sqrt(3)/2)^2 - (sqrt(10 - 2sqrt(5))/4)^2)  ...[because cos 36^circ = (sqrt(5) + 1)/4 and sin 36^circ = sqrt(10 - 2sqrt(5))/4]`

= `(sqrt(3)/4(sqrt(5) + 1))/(3/4 - ((10 - 2sqrt(5))/16)`

= `(sqrt(3)(sqrt(5) + 1))/4 xx 16/(12 - 10 + 2sqrt(5))`

= `(sqrt(3)(sqrt(5) + 1))/4 xx 16/(2(sqrt(5) + 1)`

∴  cosec 96° + cosec 384 ° = `2sqrt(3)`  ...(5)

∴ L.H.S. = cosec 48° + cosec 96° + cosec 192° + cosec 384 °

= (cosec 48°+ cosec 192°) + (cosec 96° + cosec 384°)

= `-2sqrt(3) +  2sqrt(3)` .....[By (4) and (5)]

= 0

= R.H.S.

Alternative Method :

Consider,

cosec x + cot x =  `1/sinx + cosx/sinx`

= `(1 + cosx)/sinx`

= `(2cos^2  x/2)/(2sin  x/2 * cos  x/2)`

∴ cosec x + cot x = `cot  x/2`

∴ cosec x = `cot  x/2 - cot x`  ...(1)

Replacing x by 2x, 4x, 8x in (4), we get,

cosec 2x = cot x  – cot 2x       ...(2)

cosec 4x = cot 2x – cot 4x      ...(3)

cosec 8x = cot 4x – cot 8x      ...(4)

Adding (1), (2), (3) and (4), we get

cosec x + cosec 2x + cosec 4x + cosec 8x = `cot  x/2 - cot8x`  ...(5)

By substituting x = 48° in (5), we get,

cosec 48° + cosec 96° + cosec 192° + cosec 384°

= cot 24° – cot 384°

= cot 24° – cot (360° + 24°)

= cot 24° – cot 24°  ...[∵ cot (2π + θ) = cot θ]

∴ cosec 48° + cosec 96° + cosec 192° + cosec 384° = 0

Concept: Trigonometric Functions of Allied Angels
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Balbharati Mathematics and Statistics 1 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 3 Trigonometry - 2
Miscellaneous Exercise 3 | Q II. (16) | Page 58
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×