1. ## prove identity

Prove the following identites;

$cosecAsec^{2}A = cosecA +tanAsecA$

I know you have to use these identiests; $1 + tan^{2} A = sec^{2}A$
$1+ cot^{2} A = cosec^{2}A$

2. Hello Tweety
Originally Posted by Tweety
Prove the following identites;

$cosecAsec^{2}A = cosecA +tanAsecA$

I know you have to use these identiests; $1 + tan^{2} A = sec^{2}A$
$1+ cot^{2} A = cosec^{2}A$
$\csc A \sec^2A = \csc A(1 + \tan^2 A)$

$= \csc A + \csc A \tan^2A$

$= \csc A + \frac{\sin^2 A}{\sin A\cos^2A}$

$= \csc A + \frac{\sin A}{\cos A}\cdot\frac{1}{\cos A}$

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

3. Hello, Tweety!

Sometimes it's easier to change everything to sines and cosines.

Prove: . $\csc\!A\sec^2\!\!A \:=\: \csc\! A +\tan\! A\sec\! A$

On the right side, we have: . $\frac{1}{\sin\! A} + \frac{\sin\! A}{\cos\! A}\!\cdot\!\frac{1}{\cos\! A} \;\;=\;\;\frac{1}{\sin\! A} + \frac{\sin\! A}{\cos^2\!\!A}$

Get a common denominator: . $\frac{1}{\sin\! A}\cdot{\color{blue}\frac{\cos^2\!\!A}{\cos^2\!\!A }} + \frac{\sin\!A}{\cos^2\!\!A}\cdot{\color{blue}\frac {\sin\!A}{\sin\!A}} \;\;=\;\;\frac{\cos^2\!\!A}{\sin\!A\cos^2\!\!A} + \frac{\sin^2\!\!A}{\sin\!A\cos^2\!\!A}$

. . $= \;\frac{\overbrace{\cos^2\!\!A + \sin^2\!\!A}^{\text{This is 1}}}{\sin\!A\cos^2\!\!A} \;=\;\frac{1}{\sin\!A\cos^2\!\!A}\;=\;\frac{1}{\si n\!A}\cdot\frac{1}{\cos^2\!\!A} \;=\;\csc\!A\sec^2\!\!A$