Question

#Quality Question
@Trigonometry

If θ is Acute angle and 3sinθ = 4cosθ then Find out the value of 4sin2θ - 3cos2θ + 2.
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Answer 1

Answer:

༒ Question ➽

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

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༒ Given ➽

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

___________________________

༒ 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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Answer 2

Answer:

༒ Question ➽

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

___________________________

༒ Given ➽

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

___________________________

༒ To Find ➽

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

___________________________

༒ 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,

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

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

___________________________

༒ 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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