Prove that \[ \frac{\cot x-\tan x}{\cot x+\tan x}=\sin x \].

1.	Prove that \[ \frac{\cot x-\tan x}{\cot x+\tan x}=\sin x \].
Answer» LHS = {cosx / sinx - sinx / cosx} / {cosx / sinx + sinx / cosx} LHS = {cos²x - sin²x / sinxcosx} / {cos²x + sin²x / sinxcosx} LHS = cos²x - sin²x = cos2x... (cos²x + sin²x = 1) LHS = cos2x = sinx = RHS. Hence, proved. \(\frac{cotx-tanx}{cotx+tanx}=\cfrac{\frac{cosx}{sin x}-\frac{sin x}{cos x}}{\frac{cosx}{sinx}+\frac{sinx}{cosx}}\) \(=\cfrac{\frac{cos^2x-sin^2x}{sinxcosx}}{\frac{cos^2x+sin^2x}{sinxcosx}}\) = \(\frac{cos^2x-sin^2x}{sin^2x+cos^2x}\) \(= cos^2x-sin^2x\) (\(\because\) sin²x + cos²x = 1) = cos 2x (\(\because\) cos 2x = cos²x - sin²x) \(\therefore\) \(\frac{cotx-tanx}{cotx+tanx}=cos2x\neq sinx\)

Answer»

LHS = {cosx / sinx - sinx / cosx} / {cosx / sinx + sinx / cosx}

LHS = {cos²x - sin²x / sinxcosx} / {cos²x + sin²x / sinxcosx}

LHS = cos²x - sin²x = cos2x... (cos²x + sin²x = 1)

LHS = cos2x = sinx = RHS.

Hence, proved.

\(\frac{cotx-tanx}{cotx+tanx}=\cfrac{\frac{cosx}{sin x}-\frac{sin x}{cos x}}{\frac{cosx}{sinx}+\frac{sinx}{cosx}}\)

\(=\cfrac{\frac{cos^2x-sin^2x}{sinxcosx}}{\frac{cos^2x+sin^2x}{sinxcosx}}\) = \(\frac{cos^2x-sin^2x}{sin^2x+cos^2x}\)

\(= cos^2x-sin^2x\) (\(\because\) sin²x + cos²x = 1)

= cos 2x (\(\because\) cos 2x = cos²x - sin²x)

\(\therefore\) \(\frac{cotx-tanx}{cotx+tanx}=cos2x\neq sinx\)

Discussion

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