Question

If cos x = -1÷2 - and π<x< Зπ÷2 find the value of 4tan²x - 3 Cosec²x​

Answer 1

❥GIVEN

$\sf \cos x = \frac{ - 1}{2} \: and \: \pi < x < \frac{3 \pi}{2}$

❥FIND

$\sf4 { \tan }^{2} x - 3 { \cosec }^{2} - x$

❥SOLUTION

Now, f1st we have to see that x lies in which quadrant.

in this we need a diagram ( see in attachment )

we know,

$\sf { \pi}^{e} = 80 \degree$

$\sf put \: this \: value \: in \: \pi < x < \frac{3 \pi}{2}$

so,

$\sf180 \degree < x < \frac{3 \times 180 \degree}{2}$

$\sf180 \degree < x < \frac{270 \degree}{2}$

x is lies between 180° to 270° in 3rd quadrant.

we know that,

in 1st quadrant; all trigonometric ratios are positive.

In 2nd quadrant; only sin and cosine are positive.

In 3rd quadrant; only tan and cot are positive.

In 4th quadrant; only sec and cos are positive.

$\sf \cos x ,$

we know that,

$\sf { \cos }^{2} x + { \sin }^{2} x = 1$

$\sf\sin x = + - \sqrt{1 - { \cos}^{2} x}$

$\sf\sin x = - \sqrt{1 - { \cos}^{2} x}$

$\sf\sin x = - \sqrt{ {1 - { ( \frac{ { - 1} }{2}) }^{2} } }$

$\sf\sin x = - {1 - \frac{ \sqrt{1} }{4} }$

$\sf\sin x = - { \frac{ \sqrt{3} }{2} }$

$\sf \therefore \cosec x = \frac{1}{ \sin x } ,$

$\sf \cosec x = \frac{1}{ \frac{ - \sqrt{3} }{2} } = \frac{ - 2}{ \sqrt{3} }$

$\sf \tan x = \frac{ \sin x }{ \cos x } \: by \: using \: this \: formula \\ \sf we \: can \: find \: out \: \tan x$

$\sf \tan x = \frac{ \frac{ - \sqrt{3} }{2} }{ \frac{ - 1}{2} } \\ \sf = > \tan x = \frac{ - \sqrt{3} }{2} \times - 2 = \sqrt{3}$

$\sf4 { \tan }^{2} x - 3 { \cosec }^{2} - x$

put value of tan x and cosec x in the above eq.

$\sf = > 4( \sqrt{3} {)}^{2} - 3( \frac{ - 2}{ \sqrt{3} } {)}^{2} \\ \sf = > 4 \times 3 - 3 \times \frac{4}{3} \\ \sf = > 12 - 4 = 8$

$\sf{ \green{ Hence, \: \boxed{ \sf4 { \tan }^{2}x - 3 { \cosec}^{2}x = 8. } }}$