2tanx/1-tan^2x = 2/cotx-tanx
csc^2x/cox^2x = sec^2x + csc^2
(sinx+cosx)(tanx+cotx) = secx+cscx
and
1-sinx/1+sinx = (secx - tanx)^2
Thanks alot
2tanx/1-tan^2x = 2/cotx-tanx
2tanx/1-tan^2x = tan2x = sin2x/cos2x
= 2sinx*cosx/ (cos^2x - sin^2x). Divide numerator and denomenator by sinx*cosx, you get
= 2/(cotx - tanx)
csc^2x/cox^2x = sec^2x + csc^2
sec^2x + csc^2
= 1/cos^2x + 1/sin^2x
= (sin^2x + cos^2x)/ sin^2x*cos^2x
= 1/(sin^2x*cos^2x)
= csc^2x/cos^2x
Hello, cannon!
Multiply the left side by: .$\displaystyle \frac{1-\sin x}{1-\sin x}$$\displaystyle \frac{1-\sin x}{1+\sin x} \:=\: (\sec x - \tan x)^2$
. . $\displaystyle \frac{1-\sin x}{1 + \sin x}\cdot{\color{blue}\frac{1-\sin x}{1-\sin x}} \;\;=\;\;\frac{(1-\sin x)^2}{1-\sin^2\!x} \;\;=\;\; \frac{(1-\sin x)^2}{\cos^2\!x} $ . $\displaystyle = \;\left(\frac{1-\sin x}{\cos x}\right)^2 \;=\;\left(\frac{1}{\cos x} - \frac{\sin x}{\cos x}\right)^2 \;=\;(\sec x - \tan x)^2$
Hello cannon$\displaystyle \frac{2\sin x\cos x}{\cos^2x-\sin^2x}=\frac{\dfrac{2\sin x\cos x}{\color{red}\sin x \cos x}}{\dfrac{\cos^2x-\sin^2x}{\color{red}\sin x\cos x}}$
$\displaystyle =\frac{2}{\dfrac{\cos^2x}{\sin x\cos x}-\dfrac{\sin^2x}{\sin x\cos x}}=\frac{2}{\dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x}}=\frac{2}{\cot x - \tan x}$
$\displaystyle (\sin x +\cos x)(\tan x + \cot x) = (\sin x +\cos x)\Big(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\Big)$
$\displaystyle = (\sin x +\cos x)\Big(\frac{\sin^2 x+\cos^2 x}{\cos x\sin x}\Big)$
$\displaystyle =\frac{\sin x+\cos x} {\cos x\sin x}$
$\displaystyle =\sec x + \csc x$
Grandad
Hello cannonI was picking up on the reply by sa-ri-ga-ma:
and your query As far as your next post is concerned:this thread is now too complicated! Please have another attempt at these questions - they are very similar to the ones we've shown you already. If you still can't do them, re-post them and show us your attempts in a new thread. (See Rule #14).
Grandad