Hello, <br><br>Thanks for all the help I have received on this forum<br><br>Can you please help with this questions:<br><br>Prove the identity of:<br><br>1. tan x -1 = sin^2 x - cos^2x/ sin x cos x + cos^2x<br><br>for question 1 this what i have done so far:<br><br>sin x/cos x - 1 = 1- cos^2 x - 1 - sin^2 x/sin x cosx + 1 - sin^2 x<br><br><br>2. sin x + sin x cot^2 x = csc x<br><br><br>For question 2 this is what i have done so far:<br><br>sin x + sin x 1/tan x = csc<br><br>sin x(sin x/sin x + (sin x)(1/sin x/cos x) = csc x<br><br>I have tried different methods to obtain the desired answer but I always get stuck. <br>Can you guide me through this? Thanks<br><br>yfelix45
trigonometry: proving identities
- Thread starter yfelix45
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I cannot decipher your post. May be you should write the problem in standard ASCII format.
Please review your post and use grouping symbols (parentheses) following standard hierchy of operations.
for example, if you want to write:
\(\displaystyle f(x) \ = \dfrac{1-cos(x)}{1-sin(x)}\)
in ASCII text, you should write:
f(x) = [1-cos(x)]/[1-sin(x)]
Hello, yfelix45!
The right side is: .\(\displaystyle \dfrac{(\sin x - \cos x)(\sin x + \cos x)}{\cos x(\sin x + \cos x)} \;=\;\dfrac{\sin x - \cos x}{\cos x} \;=\; \dfrac{\sin x}{\cos x} - \dfrac{\cos x}{\cos x} \;=\;\tan x - 1\)
The left side is: .\(\displaystyle \sin x(1 + \cot^2\!x) \;=\;\sin x\csc^2\!x \;=\;\dfrac{1}{\csc x}\cdot \csc^2x \;=\;\csc x \)
The right side is: .\(\displaystyle \dfrac{(\sin x - \cos x)(\sin x + \cos x)}{\cos x(\sin x + \cos x)} \;=\;\dfrac{\sin x - \cos x}{\cos x} \;=\; \dfrac{\sin x}{\cos x} - \dfrac{\cos x}{\cos x} \;=\;\tan x - 1\)
The left side is: .\(\displaystyle \sin x(1 + \cot^2\!x) \;=\;\sin x\csc^2\!x \;=\;\dfrac{1}{\csc x}\cdot \csc^2x \;=\;\csc x \)
