# Math Help - Need proving some trig identities!!!!!

1. ## Need proving some trig identities!!!!!

I'd greatly appreciate is somebody proved the following true.

1. 1-sinx/1+sinx = (secx-tanx)^2

2. sinx+1/cosx+cotx = tanx

3. sin^2x-tan^2x/1-sec^2x = sin^2x

2. $(\sec x-\tan x)^2=\left(\frac{1}{\cos x}-\frac{\sin x}{\cos x}\right)^2=\frac{(1-\sin x)^2}{\cos ^2x}=\frac{(1-\sin x)^2}{1-\sin ^2x}=\frac{1-\sin x}{1+\sin x}$

$\frac{\sin x+1}{\cos x+\cot x}=\frac{\sin x+1}{\cos x+\frac{\cos x}{\sin x}}=\frac{\sin x(\sin x+1)}{\cos x(\sin x+1)}=\tan x$

$\frac{\sin ^2x-\tan ^2x}{1-\sec ^2x}=\frac{\sin ^2x-\frac{\sin ^2x}{\cos ^2x}}{1-\frac{1}{\cos ^2x}}=\frac{\sin ^2x(\cos ^2x-1)}{\cos ^2x-1}=\sin ^2x$

3. ## problems two and three

For second and third problems, how does the second step turn into the third step.

4. Originally Posted by fam600
2. sinx+1/cosx+cotx = tanx
For the love of God, PLEASE use parenthesis! I wasted more than 10 minutes struggling with this before I understood what you meant. Your question should be typed as
(sinx+1)/(cosx+cotx) = tanx

If you don't understand the confusion, then please review order of operations.

Okay.
$\frac{sin(x) + 1}{cos(x) + cot(x)} = tan(x)$

$\frac{sin(x) + 1}{cos(x) + \frac{cos(x)}{sin(x)} }= tan(x)$

$\frac{sin(x) + 1}{cos(x) + \frac{cos(x)}{sin(x)} } \cdot \frac{sin(x)}{sin(x)} = tan(x)$

$\frac{sin^2(x) + sin(x)}{sin(x)cos(x) + cos(x)} = tan(x)$

$\frac{sin(x)(sin(x) + 1)}{cos(x)(sin(x) + 1)} = tan(x)$

$\frac{sin(x)}{cos(x)} = tan(x)$

$tan(x) = tan(x)$

-Dan

5. Originally Posted by fam600
3. sin^2x-tan^2x/1-sec^2x = sin^2x
Let me guess:
(sin^2x-tan^2x)/(1-sec^2x) = sin^2x

$\frac{sin^2(x) - tan^2(x)}{1 - sec^2(x)} = sin^2(x)$

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

$\frac{sin^2(x) - \frac{sin^2(x)}{cos^2(x)}}{1 - \frac{1}{cos^2(x)}} \cdot \frac{cos^2(x)}{cos^2(x)} = sin^2(x)$

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

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

$sin^2(x) = sin^2(x)$

-Dan

6. Hello, fam600!

Different approaches . . .

$1)\;\frac{1-\sin x}{1+\sin x} \:=\: (\sec x-\tan x)^2$

Multiply top and bottom by $(1-\sin x)$

. . $\frac{1-\sin x}{1 +\sin x}\cdot\frac{1 -\sin x}{1-\sin x} \;\;=\;\;\frac{1-2\sin x+\sin^2x}{1-\sin^2x} \;\;=\;\;\frac{1-2\sin x+\sin^2x}{\cos^2x}$

. . $= \;\;\frac{1}{\cos^2x} - \frac{2\sin x}{\cos^2x} + \frac{\sin^2x}{\cos^2x} \;\;=\;\;\frac{1}{\cos^2x} -2\left(\frac{1}{\cos x}\right)\left(\frac{\sin x}{\cos x}\right) + \left(\frac{\sin x}{\cos x}\right)^2$

. . $=\;\;\sec^2x - 2\sec x\tan x + \tan^2x \;\;=\;\;(\sec x - \tan x)^2$

$2)\;\frac{\sin x+1}{\cos x+\cot x} \:=\:\tan x$

We have: . $\frac{\sin x + 1}{\cos x + \frac{\cos x}{\sin x}}\;=\;\frac{\sin x + 1}{\cos x\left(1 + \frac{1}{\sin x}\right)} \;=\;\frac{\sin x+1}{\cos x\left(\frac{\sin x + 1}{\sin x}\right)} \;=\;\frac{1}{\frac{\cos x}{\sin x}} \;=\;\frac{\sin x}{\cos x} \;=\;\tan x$

$3)\;\frac{\sin^2x-\tan^2x}{1-\sec^2x} \:=\:\sin^2x$

We have: . $\frac{\sin^2x - \frac{\sin^2x}{\cos^2x}}{1 - \sec^2x} \;=\;\frac{\sin^2x\left(1 - \frac{1}{\cos^2x}\right)}{1-\sec^2x} \;=\;\frac{\sin^2x(1 - \sec^2x)}{1-\sec^2x} \;=\;\sin^2x$

7. Thanks alot everyone. Sorobon, those different approaches especially helped. Its been a year since my last algebra course and so i've forgotten some of it, thats why i've been having such a hard time with these.