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

Sum

Prove the following:

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

#### 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
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#### 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