How is this wrong solution?

dollyayesha2345

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If sinx=cosxsin x=cos x then find the value of 2tanxcos2x2tanx-cos^2x.

My attempt:
sin xcos x=cos xcos x\frac{sin\ x}{cos\ x}=\frac{cos\ x}{cos\ x}tanx=1tan x =1x=tan1x=tan^{-1}x=45°x=45°Substituting x=45°x=45° in 2tanxcos2x2tanx-cos^2x2tan45°cos2(45°)2tan45°-cos^2(45°)2tan45°cos90°2tan45°-cos90° (I received a feedback saying "wrong solution" on this step exactly)
=2(1)0=2(1)-0=2=2
 
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Where's the solution? You should know how the forum works by now. Would you please show your work and tell us why you're stuck?
 
2tan45°−cos2(45°)2tan45°−cos90°2tan45°-cos90°2tan45°−cos90°
The error you made is here.
cos2(45°)cos(90°)cos^2(45\degree)\neq cos(90\degree)cos2(45°)=cos(45°)cos(45°)cos^2(45\degree) = cos(45\degree)*cos(45\degree)
 
The error you made is here.
cos2(45°)cos(90°)cos^2(45\degree)\neq cos(90\degree)cos2(45°)=cos(45°)cos(45°)cos^2(45\degree) = cos(45\degree)*cos(45\degree)
So I guess it should be like this:
2tan45°cos2(45°)2tan45°−cos^2 (45°)=2(1)12=2(1)-\frac{1}{2}=32=\frac{3}{2}Right?
 
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So I guess it should be like this:
2tan45°cos2(45°)2tan45°−cos 2 (45°)=2(1)×=2(1)\times2(1)×122(1)\times\frac{1}{2}=32=\frac{3}{2}Right?
Not quite. Think cos2(45°)cos^2(45\degree) similar to x2=xxx^2=x*x. You evaluate cos(45°)cos(45\degree) first, then square it. That's what the notation means. Another way you can write it as: cos2(45°)=[cos(45°)]2cos^2(45\degree) = [cos(45\degree)]^2
 
Not quite. Think cos2(45°)cos^2(45\degree) similar to x2=xxx^2=x*x. You evaluate cos(45°)cos(45\degree) first, then square it.
cos245=(22)2cos^245=(\frac{\sqrt2}{2})^2(22)2=12(\frac{\sqrt2}{2})^2=\frac{1}{2}
 
tanx=1tan x =1x=tan1x=tan^{-1}x=45°x=45°
tan-1 is a function, that is for any input you'll never get back more than one answer.

tan x =1 and x=tan-1(1) are NOT the same!

tan x = 1 has infinitely many solutions where as x=tan-1(1) has just one solution.
There is a solution in quadrant I, which you found out to be x=π/4. But that is not complete. x can equal π/4±n∗2π, when n is any integer.
Now is there another quadrant where tan x = 1? What angle would that be?
 
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