Question

sinA/1-COSA+tanA/1+cosA=2secAcosecA+cotA prove that

Answer 1

Answer on question 34067 – Math – Trigonometry

sinA/1-COSA+tanA/1+cosA=2secAcosecA+cotA prove that

Proving

$$\frac {\sin A}{1 - \cos A} + \frac {\tan A}{1 + \cos A} =$$

(in the first fraction we multiply the denominator and denominator by $(1 + \cos A)$,

in the second fraction by $(1 - \cos A) = \frac{\sin A(1 + \cos A)}{(1 - \cos A)(1 + \cos A)} + \frac{\tan A(1 - \cos A)}{(1 + \cos A)(1 - \cos A)} =$

$$= \frac {\sin A (1 + \cos A)}{1 - \cos^ {2} A} + \frac {\tan A (1 - \cos A)}{1 - \cos^ {2} A}$$

= (using the main trigonometry identity and the definition of $\tan$) =

$$= \frac {\sin A (1 + \cos A)}{\sin^ {2} A} + \frac {\sin A (1 - \cos A)}{\sin^ {2} A \cos A} = \frac {(1 + \cos A)}{\sin A} + \frac {(1 - \cos A)}{\sin A \cos A} = \frac {\cos A + \cos^ {2} A + 1 - \cos A}{\sin A \cos A} =$$ $$= \frac {\cos^ {2} A + 1}{\sin A \cos A} = \frac {1}{\sin A \cos A} + \frac {\cos^ {2} A}{\sin A \cos A}$$

= (using the definition of $\sec A$, $\csc A$ and tangent we get)

$$= \sec A \csc A + \tan A.$$

QED.

www.AssignmentExpert.com