Chapter 7.4, Problem 70AYU

establish each identity.

$\frac{{\sin }^2\theta -\tan \theta }{{\cos }^2\theta -\cot \theta }={\tan }^2\theta$

To determine

To find: The trigonometric expression $\frac{{\text{sin}}^2\theta –\tan \theta }{{\text{cos}}^2\theta –\cot \theta }={\text{ tan}}^2\theta$.

Answer to Problem 70AYU

Solution:

The trigonometric expression $\frac{{\text{sin}}^2\theta –\tan \theta }{{\text{cos}}^2\theta –\cot \theta }={\text{ tan}}^2\theta$ is established.

Explanation of Solution

Given:

The trigonometric expression $\frac{{\text{sin}}^2\theta –\tan \theta }{{\text{cos}}^2\theta –\cot \theta }={\text{ tan}}^2\theta$.

Calculation:

Left hand side

$\frac{{\text{sin}}^2\theta –\tan \theta }{{\text{cos}}^2\theta –\cot \theta }$

By quotient identity for $\tan \theta ,\cot \theta$

$\frac{{\text{sin}}^2\theta –\frac{\sin \theta }{\cos \theta }}{{\text{cos}}^2\theta –\frac{\cos \theta }{\sin \theta }}$

Taking $\sin \theta$ from numerator and $\cos \theta$ from denominator

$\frac{\sin \theta \left(\sin \theta –\frac{1}{\cos \theta }\right)}{\cos \theta \left(\cos \theta –\frac{1}{\sin \theta }\right)}$

By quotient identity

$\frac{\tan \theta .\left(\sin \theta –\frac{1}{\cos \theta }\right)}{\left(\cos \theta –\frac{1}{\sin \theta }\right)}$

Adding the quotients in the numerator and denominator

$\frac{\tan \theta .\left(\frac{\sin \theta \cos \theta -1}{\cos \theta }\right)}{\left(\frac{\cos \theta \sin \theta –1}{\sin \theta }\right)}$

Cancelling the common factor $(\sin \theta \cos \theta -1)$ and taking reciprocal

$\frac{\tan \theta .\sin \theta }{\cos \theta }$

By quotient identity

$\tan \theta \tan \theta$

${\text{tan}}^2\theta$

$\text{LHS = RHS}$