Question

Chancel opted wooden stick With______ Than phosphorus

Answer 1

Answer:

Instead of using phosphorus, Chancel elected to coat wooden stick with potassium chlorate, sulfur, sugar, rubber, and then dip that stick into the small asbestos bottle filled with sulfuric acid.

Answer 2

Explanation:

༒ Question ➽

If θ is Acute angle and 3sinθ = 4cosθ then Find out the value of 4sin2θ - 3cos2θ + 2.

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༒ Given ➽

$\large \sf3sin \theta = 4cos\theta$

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༒ To Find ➽

$\sf \large 4sin2\theta - 3cos2\theta + 2$

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༒ Solution ➽

As

$\sf3sin\theta = 4cos\theta \\ \\ \implies \sf \dfrac{sin\theta }{cos \theta } = \frac{4}{3} \\ \\ \implies \large \boxed{ \bf tan\theta = \frac{4}{3} }$

So,

$\sf {tan}^{2} \theta = \frac{16}{9} \\ \\ \sf \implies {sec}^{2} \theta = 1 + \frac{16}{9} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \{ \because\bf {sec}^{2} \theta = 1 + tan {}^{2} \theta \} \\ \\ = \sf \frac{25}{9} \\ \\ \implies \sf sec \theta = \pm \frac{5}{3} \\ \ \bf \{but \: sec\theta \ne - ve \} \: \: ( \because \theta \: is \: acute) \\ \\ \therefore \bf sec\theta = \frac{5}{3}$

Hence,

$\purple{ \large \underline{ \boxed{\bf cos\theta = \dfrac{3}{5}}}} \\ \: \: \: \: \: \: \: \{ \bf\because cos\theta = \frac{1}{sec \theta } \}$

So,

$\sf sin\theta = \sqrt{1 - {cos}^{2} \theta } \\ \\ \sf = \sqrt{1 - \frac{9}{25} } \\ \\ = \sf \sqrt{ \frac{16}{25} } \\ \\ \sf = \pm\frac{4}{5}$

As θ is Acute so sinθ will not negative

Hence,

$\large \purple{ \bf \underline{ \boxed{ \bf sin \theta = \frac{4}{5} }}}$

Now,

$\sf4sin2\theta - 3cos2\theta + 2 \\ \\ = \sf 4(2sin\theta cos\theta) - 3(2 {cos}^{2} \theta - 1) + 2 \\ \\ \sf = 8 \times \frac{4}{5} \times \frac{3}{5} -3 \bigg\{2\bigg( \frac{3}{5} \bigg) {}^{2} - 1 \bigg\} + 2 \\ \\ = \sf \frac{96}{25} + \frac{21}{25} + 2 \\ \\ \Large \purple{\bf = \frac{167}{25} }$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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༒ Alernate Solution ➽

As

$\sf3sin\theta = 4cos\theta \\ \\ \implies \sf \dfrac{sin\theta }{cos \theta } = \frac{4}{3} \\ \\ \implies \large \boxed{ \bf tan\theta = \frac{4}{3} }$

Squaring We get

$\bf {tan}^{2} \theta = \frac{16}{9}$

We Know

$\sf sin2 \theta = \dfrac{2tan \theta}{1 + {tan}^{2} \theta} \\ \\ \bf and \\ \\ \sf cos2 \theta = \frac{1 - {tan}^{2} \theta}{1 + tan {}^{2} \theta }$

Now,

$\sf4sin2\theta - 3cos2\theta + 2 \\ \\ = \sf4 \bigg( \frac{2tan\theta}{1 + tan {}^{2} \theta } \bigg) - 3 \bigg( \frac{1 - {tan}^{2} \theta}{1 + tan {}^{2} \theta } \bigg) + 2 \\ \\ \sf = 4 \bigg( \frac{2( \frac{4}{3}) }{1 + \frac{16}{9} } \bigg) - 3 \bigg( \frac{1 - \frac{16}{9} }{1 + \frac{16}{9} } \bigg) + 2 \\ \\ \sf = 4 \bigg( \dfrac{ 8/3 }{ 25 /9 } \bigg) - 3 \bigg( \frac{ - 7 /9 }{ 25 /9 } \bigg) + 2 \\ \\ \sf = 4 \bigg( \frac{24}{5} \bigg) + 3 \bigg( \frac{7}{25} \bigg) + 2 \\ \\ \sf = \frac{96}{25} + \frac{21}{25} + 2 \\ \\ \Large \bf \purple{ \bf = \frac{167}{25} }$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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