Question

prove that 1-sin^2x/1+cotx-cos^2x/1+tanx=sinxcossx

Answer 1

$\textbf{To prove:}$

$1-\dfrac{sin^2x}{1+cot\,x}-\dfrac{cos^2x}{1+tan\,x}=sinx\,cosx$

$\textbf{Solution:}$

$\text{Consider,}$

$1-\dfrac{sin^2x}{1+cot\,x}-\dfrac{cos^2x}{1+tan\,x}$

$=1-\dfrac{sin^2x}{1+\frac{cos\,x}{sin\,x}}-\dfrac{cos^2x}{1+\frac{sin\,x}{cos\,x}}$

$=1-\dfrac{sin^3x}{sin\,x+cos\,x}-\dfrac{cos^3x}{sin\,x+cos\,x}$

$=\dfrac{(sin\x+cos\,x)-sin^3x-cos^3x}{sin\,x+cos\,x}$

$=\dfrac{(sin\x+cos\,x)-(sin^3x+cos^3x)}{sin\,x+cos\,x}$

$=\dfrac{(sin\x+cos\,x)-(sin\,x+cos\,x)(sin^2x-sin\,x\,cos\,x+cos^2x)}{sin\,x+cos\,x}$

$=\dfrac{(sin\x+cos\,x)-(sin\,x+cos\,x)(1-sin\,x\,cos\,x)}{sin\,x+cos\,x}$

$=(sin\x+cos\,x)(\dfrac{1-(1-sin\,x\,cos\,x)}{sin\,x+cos\,x})$

$=1-(1-sin\,x\,cos\,x)$

$=1-1+sin\,x\,cos\,x$

$=sin\,x\,cos\,x$

$\implies\boxed{\bf\,1-\dfrac{sin^2x}{1+cot\,x}-\dfrac{cos^2x}{1+tan\,x}=sinx\,cosx}$

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