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how to evaluate the lim x-pi/4 (1-tan x)sec 2x?

Guest May 26, 2015

Best Answer 

 #1
avatar+26329 
+15

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

 

tan(pi/4) = 1  and cos(pi/4) = 1/√2 so:

 

$$\frac{1}{(1+\tan{\frac{\pi}{4}})\cos^2{\frac{\pi}{4}}}=\frac{1}{2\times \frac{1}{2}}=1$$

 

.

Alan  May 26, 2015
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2+0 Answers

 #1
avatar+26329 
+15
Best Answer

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

 

tan(pi/4) = 1  and cos(pi/4) = 1/√2 so:

 

$$\frac{1}{(1+\tan{\frac{\pi}{4}})\cos^2{\frac{\pi}{4}}}=\frac{1}{2\times \frac{1}{2}}=1$$

 

.

Alan  May 26, 2015
 #2
avatar+91038 
+10

Thanks Alan,

I doubt I would have got that on my own EVEN if I had been a good enother forensic mathematician to work out the question in the first place.   LOL

 

I put this into the Great Answers to Learn from Sticky Thread :)

Melody  May 27, 2015

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