Question

1. If tan x = -3/4 and x is in the second quadrant, then what is the value of
sin x. cos x?​

Answer 1

▪Given :-

$\large \pink{ \bf tan \: x = \dfrac{ - 3}{ \: 4} } \\ \bf and \\ \pink{ \bf x \in2nd \: quadrant}$

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▪To Find :-

$\mathfrak{\text{T}he \: value \: of \: } \\ \large\red{ \bf sin \: x.cos \: x}$

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▪Solution :-

$\sf tan \: x = - \dfrac{3}{4} \\ \\ \sf {tan}^{2} x = \frac{9}{16} \\ \\ \sf \sec {}^{2} x = 1 + \frac{9}{16} = \frac{25}{16}$

$\sf sec \: x = \sqrt{ \frac{25}{16 } } \\ \\ \sf sec \: x = \pm \frac{5}{4}$

But sec x Can't be positive As x lies in 2nd quadrant

Hence,

$\sf sec \: x = - \dfrac{5}{4} \\ \\ \implies \Large \purple{ \bf \underline{ \boxed{ \bf cos \: x = - \frac{4}{5} }}}$

As, $\bf {sin}^{2} x + cos {}^{2} x = 1$

So,

$\sf sin \: x = \sqrt{1 - {cos}^{2} \: x} \\ \\ = \sf \sqrt{1 - \frac{16}{25} } \\ \\ = \sqrt{ \frac{9}{25} } \\ \\ \sf sin \: x = \pm \frac{3}{5}$

As x lies in 2nd quadrant So sin x Can't be Negative

$\implies\Large \purple{ \bf \underline{ \boxed{ \bf sin \: x = \frac{3}{5} }}}$

So,

$\sf sin \: x.cos \: x = \frac{3}{5} \times ( - \frac{4}{5} ) \\ \\ \implies\Large \purple{ \bf \underline{ \boxed{ \bf sin \: x. cos \: x = - \frac{12}{25} }}}$

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

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Answer 2

Given :-

If tan x = -3/4 and x is in the second quadrant,

To Find :-

sin x × cos x

Solution :-

On squaring both sides

(tan² x) = (-3/4)²

tan²x = (-3²/4²)

tan²x = 9/16

Now

We know that

sec² x = 1 + tan²x

sec²x = 1 + 9/16

sec²x = 16 + 9/16

sec²x = 25/16

sec x = √(25/16)

sec x = -5/4

Now cos x = 1/sec

cos x = 1/(-5/4)

cos x = -4/5

sin² x + cos² x = 1

sin² x + (-4/5)² = 1

sin² x + 16/25 = 1

sin² x = 1 - 16/25

sin x = √(25 - 16/25)

sin x = √(9/25)

sin x = 3/5

Now

sin x × cos x = 3/5 × -4/5

sin x × cos x = -12/25