Question

The value of $\frac{cos^{3}20^{0}-cos^{3}70^{0}}{sin^{3}70^{0}-sin^{3}20^{0}}$ is
(a)$\frac{1}{2}$
(b)$\frac{1}{√2}$
(c)1
(d)2

Answer 1

SOLUTION :  

The correct option is  (c) = 1

Given : cos³ 20° - cos³ 70°/ sin³ 70° - sin³ 20°

cos³ 20° - cos³ 70°/ sin³ 70° - sin³ 20°

= [(cos 20° - cos 70°) (cos² 20° + cos² 70° + cos 20° cos 70°)] / [(sin 70° - sin 20°)(sin² 70° + sin² 20°+ sin 70° sin 20°

[(a³ - b³) = (a -  b) (a² + b² + ab]

= [(cos (90° - 70°) - cos (90° - 20°)) (cos² (90° - 70°)+ cos² 70° + cos 20° cos 70°] / [(sin 70° - sin 20°)(sin² 70° + sin²(90° -  70°)) + sin (90° - 20° ) sin (90° - 70°)]

= (sin 70° - sin 20°)(sin² 70° + cos² 70° + cos 20° cos 70°] / [(sin 70° - sin 20°)(sin² 70° + cos ²  70°)) + cos 20° cos 70°)]

[cos (90 - θ) = sin θ , sin (90° -θ ) = cos θ]

= (1 + cos 20° cos 70°) / (1 + cos 20° cos 70°)

[sin² θ + cos²  θ = 1]

= 1

cos³ 20° - cos³ 70°/ sin³ 70° - sin³ 20° = 1

Hence, the value of cos³ 20° - cos³ 70°/ sin³ 70° - sin³ 20° is 1 .

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Answer:

Option(C)

Step-by-step explanation:

Given Equation is : [cos³20 - cos³70]/[sin³70 - sin³20]

= [cos³(90 - 70) - cos³70] / [sin³ 70 - sin³(90 - 70)]

∴ cos(90 - θ) = sin θ and sin(90 - θ) = cos θ.

= [sin³ 70 - cos³ 70] / [sin³ 70 - cos³ 70]

= 1.

Hope it helps!