Question

Prove That : cos²A / 1 - tan A + sin³A / sinA - cos A = 1 + sin A . cos A​

Answer 1

$\tt \huge {\boxed {\mathbb {QUESTION}}}$

$\tt\dfrac{{cos}^{2}\: \theta}{1-tan\: \theta} +\dfrac{{sin}^{3}\: \theta}{sin\: \theta-cos\: \theta} =1+sin\: \theta.cos\: \theta$

$\tt \huge {\boxed {\mathbb {ANSWER}}}$

$\tt\implies \dfrac{{cos}^{2}\: \theta}{1-tan\: \theta} +\dfrac{{sin}^{3}\: \theta}{sin\: \theta-cos\: \theta} =1+sin\: \theta.cos\: \theta$

$\tt LHS$

$\tt\implies \dfrac{{cos}^{2}\: \theta}{1-tan\: \theta} +\dfrac{{sin}^{3}\: \theta}{sin\: \theta-cos\: \theta}$

$\tt\implies \dfrac{{cos}^{2}\: \theta}{1-\frac{sin\: \theta}{cos\: \theta}} - \frac{{sin}^{3}\: \theta}{sin\: \theta-cos\: \theta}$

$\tt\implies \dfrac{{cos}^{2}\: \theta}{\frac{cos\: \theta-sin\: \theta}{cos\: \theta}} - \frac{{sin}^{3}\: \theta}{sin\: \theta-cos\: \theta}$

$\tt\implies \dfrac{{cos}^{2}\: \theta}{cos\: \theta-sin\: \theta} \times cos\: \theta - \frac{{sin}^{3}\: \theta}{sin\: \theta-cos\: \theta}$

$\tt\implies \dfrac{{cos}^{3}\: \theta}{sin\: \theta-cos\: \theta} - \frac{{sin}^{3}\: \theta}{sin\: \theta-cos\: \theta}$

$\tt\implies \dfrac{{cos}^{3}\: \theta-{sin}^{3}\: \theta}{sin\: \theta-cos\: \theta}$

$\tt \implies \dfrac {\cancel {(cos\: \theta - sin\: \theta)} ({cos}^{2}\: \theta+cos \theta.sin\: \theta+{sin}^{2}\: \theta}{\cancel {(cos\: \theta - sin\: \theta)}}$

$\tt \implies ({cos}^{2}\:\theta+{sin}^{2}\:\theta+sin\:\theta.cos\:\theta)$

$\tt\implies 1+sin\:\theta.cos\:\theta$

$\tt RHS$

$\tt Used\:Formulas$

$\tt{sin}^{2}\:\theta+{cos}^{2}\:\theta=1$

$\tt{a}^{3}-{b}^{3}=(a-b)({a}^{2}+ab+{b}^{2})$

$\tt \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

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$\tt \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

$\tt Formulas$

$\tt {sin}^{2}\:\theta+{cos}^{2}\:\theta=1$

$\tt 1+{tan}^{2}\:\theta={sec}^{2}\:\theta$

$\tt 1+{cot}^{2}\:\theta={cosec}^{2}\:\theta$

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