dollyayesha2345
Junior Member
- Joined
- Oct 28, 2021
- Messages
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If [math]sin x=cos x[/math] then find the value of [math]2tanx-cos^2x[/math].
My attempt:
[math]\frac{sin\ x}{cos\ x}=\frac{cos\ x}{cos\ x}[/math][math]tan x =1[/math][math]x=tan^{-1}[/math][math]x=45°[/math]Substituting [math]x=45°[/math] in [math]2tanx-cos^2x[/math][math]2tan45°-cos^2(45°)[/math][math]2tan45°-cos90°[/math] (I received a feedback saying "wrong solution" on this step exactly)
[math]=2(1)-0[/math][math]=2[/math]
My attempt:
[math]\frac{sin\ x}{cos\ x}=\frac{cos\ x}{cos\ x}[/math][math]tan x =1[/math][math]x=tan^{-1}[/math][math]x=45°[/math]Substituting [math]x=45°[/math] in [math]2tanx-cos^2x[/math][math]2tan45°-cos^2(45°)[/math][math]2tan45°-cos90°[/math] (I received a feedback saying "wrong solution" on this step exactly)
[math]=2(1)-0[/math][math]=2[/math]
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