Question

cos cube 20 degree minus cos cube 70 degree by sin cube 70 degree minus sin Cube 20 degree

Answer 1

Answer:

Step-by-step explanation:

HERE IT IS...

Answer 2

The vaue of  $\dfrac{\cos^3 20-\cos^3 70}{\sin^3 20-\sin^3 70}$ is equal to one(1).

Step-by-step explanation:

We have,

$\dfrac{\cos^3 20-\cos^3 70}{\sin^3 20-\sin^3 70}$

To find, $\dfrac{\cos^3 20-\cos^3 70}{\sin^3 20-\sin^3 70}=?$

∴ $\dfrac{\cos^3 20-\cos^3 70}{\sin^3 20-\sin^3 70}$

$=\dfrac{\cos^3 (90-70)-\cos^3 (90-20)}{\sin^3 20-\sin^3 70}$

$=\dfrac{\sin^3 20-\sin^3 70}{\sin^3 20-\sin^3 70}$

[ ∵ $\cos (90-A)=\sin A$]

Numerator and denominator are same, the value is 1.

= 1

∴ $\dfrac{\cos^3 20-\cos^3 70}{\sin^3 20-\sin^3 70}=1$

Hence, the vaue of  $\dfrac{\cos^3 20-\cos^3 70}{\sin^3 20-\sin^3 70}$ is equal to one(1).