Question

Solve this trigonometry question too, please by an easy method.​

Answer 1

Answer:

proof given below

Step-by-step explanation:

Combine the fractions by making the denominators the same:

   $\dfrac{\sin \theta}{1+ \cos \theta}+\dfrac{1+ \cos \theta}{ \sin \theta}$

$=\dfrac{\sin \theta}{1+ \cos \theta} \cdot \dfrac{\sin \theta}{\sin \theta}+\dfrac{1+ \cos \theta}{ \sin \theta} \cdot \dfrac{1+ \cos \theta}{1+ \cos \theta}$

$=\dfrac{\sin \theta(\sin \theta)}{\sin \theta(1+ \cos \theta)}+\dfrac{(1+ \cos \theta)(1+ \cos \theta)}{\sin \theta(1+ \cos \theta)}$

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

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

Use the trigonometric identity $\sin^2 \theta + \cos^2 \theta=1$ :

$= \dfrac{1+1+2 \cos \theta}{\sin \theta(1+ \cos \theta)}$

$= \dfrac{2+2 \cos \theta}{\sin \theta(1+ \cos \theta)}$

Factor out 2 from the numerator:

$= \dfrac{2(1+ \cos \theta)}{\sin \theta(1+ \cos \theta)}$

Cancel the common factor:

$= \dfrac{2}{\sin \theta}$

$\textsf{Use the identity} \quad\csc \theta=\dfrac{1}{\sin \theta}:$

$=2 \csc \theta$

Hence proved.

Learn more about trigonometric identities here:

brainly.com/question/27938536