Question

evaluate 8 by cot square teta - 8 by cos square teta​

Answer 1

Step-by-step explanation:

= 8/cot^2 theta - 8/cos^2 theta

= 8sin^2theta/ cos^2theta - 8/cos^2theta

= 8sin^2theta -8/cos^2theta

= -8 (1 - sin^2theta )/cos^2theta

= -8 cos^2theta /cos^2theta

= -8

Answer 2

The value of $\dfrac{8}{\cot^2 \theta}-\dfrac{8}{\cos^2 \theta}$ = - 8

Step-by-step explanation:

We have,

$\dfrac{8}{\cot^2 \theta}-\dfrac{8}{\cos^2 \theta}$

To find, the value of $\dfrac{8}{\cot^2 \theta}-\dfrac{8}{\cos^2 \theta}$ = ?

∴ $\dfrac{8}{\cot^2 \theta}-\dfrac{8}{\cos^2 \theta}$

$=\dfrac{8}{\dfrac{\cos^2 \theta}{\sin^2 \theta} }-\dfrac{8}{\cos^2 \theta}$

Using the trigonometric identity,

$\cot \theta}=\dfrac{\cos \theta}{\sin \theta}$

$=\dfrac{8\sin^2 \theta}{\cos^2 \theta}-\dfrac{8}{\cos^2 \theta}$

= $\dfrac{8}{\cos ^2 \theta} (\sin^2 \theta-1)$

$= \dfrac{8}{\cos ^2 \theta} (-\cos ^2 \theta)$

Using the trigonometric identity,

$\sin^2 \theta-\cos^2 \theta=1$

= - 8

Thus, the value of $\dfrac{8}{\cot^2 \theta}-\dfrac{8}{\cos^2 \theta}$ = - 8