1. ## proving trigonometric identity

prove that sin A + cosA = 1
secA + tanA - 1 cosecA + cotA -1

2. Originally Posted by aman
prove that sin A + cosA = 1
secA + tanA - 1 cosecA + cotA -1
are you sure about the firs one? The $\displaystyle \sin A + \cos A = 1$?

and the second one, what should we prove?

Thanks.

3. sorry that was mistake , sinA is divided by secA + tan A - 1 and cosA is divided by cosecA + cotA -1 and they are added to get 1.

4. Originally Posted by aman
sorry that was mistake , sinA is divided by secA + tan A - 1 and cosA is divided by cosecA + cotA -1 and they are added to get 1.
$\displaystyle sin(A) + cos(A) \neq 1$

Try it with $\displaystyle A = \frac{\pi}{4} = 45^o$

However: $\displaystyle sin^2(A) + cos^2(A) = 1$

5. Originally Posted by aman
sorry that was mistake , sinA is divided by secA + tan A - 1 and cosA is divided by cosecA + cotA -1 and they are added to get 1.

This is not an identity

try $\displaystyle \frac{\pi}{4}$

$\displaystyle \sin\left( \frac{\pi}{4}\right)+\cos\left( \frac{\pi}{4}\right) =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2} \ne 1$

6. TES, you've quoted the correct version of the question, but you've taken into account the very first post lol.

We have to prove that :
$\displaystyle S=\frac{\sin(a)}{\sec(a)+\tan(a)-1}+\frac{\cos(a)}{\csc(a)+\cot(a)-1}=1$

Recall that $\displaystyle \cot=\frac{\cos}{\sin}=\frac1\tan$
$\displaystyle \sec=\frac1\cos$
$\displaystyle \csc=\frac1\sin$

$\displaystyle S=\frac{\sin(a)}{\sec(a)+\tan(a)-1}=\frac{\sin(a)}{\sec(a)+\tan(a)-1}\cdot\frac{\cot(a)}{\cot(a)}=\frac{\cos(a)}{\csc (a)+1-\cot(a)}$

Hey ! But this is very similar to $\displaystyle \frac{\cos(a)}{\csc(a)+\cot(a)-1}$

$\displaystyle \cos(a)\left(\frac{1}{\csc(a)+1-\cot(a)}+\frac{1}{\csc(a)+\cot(a)-1}\right)$

$\displaystyle =\cos(a)\left(\frac{\csc(a)+\cot(a)-1+\csc(a)+1-\cot(a)}{(\csc(a)+1-\cot(a))(\csc(a)+\cot(a)-1)}\right)$

$\displaystyle =\cos(a)\left(\frac{2\csc(a)}{(p-q)(p+q)}\right)$

where $\displaystyle p=\csc(a)$ and $\displaystyle q=1-\cot(a)$

\displaystyle \begin{aligned} (p-q)(p+q) &=p^2-q^2 \\ &=\csc^2(a)-(1-\cot(a))^2 \\ &=\csc^2(a)-\cot^2(a)+2\cot(a)-1 \end{aligned}

But it can be very easily proved (or known) that $\displaystyle \csc^2(a)-\cot^2(a)=1$

Hence $\displaystyle (p-q)(p+q)=2\cot(a)$

And finally, we have

$\displaystyle \frac{\sin(a)}{\sec(a)+\tan(a)-1}+\frac{\cos(a)}{\csc(a)+\cot(a)-1}=\cos(a)\cdot\frac{2\csc(a)}{2\cot(a)}$

But note that $\displaystyle \cos\cdot\csc=\cos\cdot\frac{1}{\sin}=\cot$

Hence $\displaystyle \boxed{\frac{\sin(a)}{\sec(a)+\tan(a)-1}+\frac{\cos(a)}{\csc(a)+\cot(a)-1}=1}$

Note : I've been doing all this while I'm not used at all to csc, cot, sec. What I used for simplifications is their definition in terms of cos, sin, tan. So the main difficulty for understanding is that I've been messy in my message and that I skipped some details lol