Question

please give solution step by step

according to 10th class​

Answer 1

$\large\bf\underline {Question:-}$

Prove that $\rm \: \frac{sin \theta - 2sin {}^{3} \theta }{2cos {}^{} \theta - cos \theta } = tan \theta$

$\large\bf\underline {To \: prove:-}$

$\rm \: \frac{sin \theta - 2sin {}^{3} \theta }{2cos {}^{} \theta - cos \theta } = tan \theta$

$\huge\bf\underline{Solution:-}$

$\rm \: \frac{sin \theta - 2sin {}^{3} \theta }{2cos {}^{} \theta - cos \theta } = tan \theta$

LHS :-

$\mapsto \rm \: \frac{sin \theta - 2sin {}^{3} \theta }{2cos {}^{3} \theta - cos \theta} \\ \\ \mapsto \rm \: \frac{sin \theta(1 - 2 {sin}^{2} \theta) }{cos \theta(2cos {}^{2} \theta - 1)} \\ \\ \mapsto \boxed{ \bf \: sin {}^{2} \theta = 1 - {cos}^{2} \theta } \\\\\mapsto \boxed{{ \bf\frac{sin\theta}{cos\theta}= tan\theta} } \\ \\ \mapsto \rm \: tan \theta \frac{(1 - 2(1 - co {s}^{2} \theta)) }{(2 {cos}^{2} \theta - 1)} \\ \\\rm \mapsto \: \tan \frac{(1 - 2 + 2 {cos}^{2} \theta)}{(2 {cos}^{2} \theta - 1) } \\ \\ \rm \mapsto \: \tan \frac{(2cos {}^{2} \theta - 1) }{ (2 {cos}^{2} \theta - 1) }\\ \\ \mapsto \rm \: tan \theta \frac{ \cancel{(2 {cos}^{2} \theta - 1)} }{ \cancel{(2 {cos}^{2} \theta - 1)} } \\ \\ \bf= tan \theta$

LHS = RHS

Hence, proved.

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Some important formulas:-

★ tan∅ = sin∅/cos∅

★ cot ∅ = cos∅/sin∅

★ tan∅.cot∅ = 1

★ sin²∅ +cos²∅ = 1

★ 1+ tan²∅ = sec²∅

★ 1 +cot²∅ = cosec²∅

Answer 2

Answer:

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