Question

1/sectheta+tantheta = 1-sin theta /costheta

Answer 1

Answer:

this is your solution for the following information

Answer 2

It is proved that $\frac{1}{\sec \theta +\tan \theta }=\frac{1-\sin \theta }{\cos \theta }$

Step-by-step explanation:

Taking the LHS of equation, i.e., $\frac{1}{\sec \theta +\tan \theta }</p><p>[tex]\Rightarrow \frac{1}{\sec \theta +\tan \theta }$

$\Rightarrow \frac{1}{\frac{1}{\cos\theta } +\frac{\sin \theta }{\cos \theta }}$

$\Rightarrow \frac{1}{\frac{1 + \sin \theta}{\cos\theta } }$

$\Rightarrow \frac{\cos\theta }{1 + \sin \theta}$

By compendo and dividendo ,multiply numerator and denominator by  ${1 - \sin \theta}$

$\Rightarrow \frac{\cos\theta \times\left ( 1 - \sin \theta \right ) }{\cos^{2} \theta}$

$\Rightarrow \frac{\left ( 1 - \sin \theta \right ) }{\cos \theta}[\tex]</strong></p><p><strong>It is proved that [tex]\frac{1}{\sec \theta +\tan \theta }=\frac{1-\sin \theta }{\cos \theta }$