Math, asked by Eshwank, 7 months ago

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

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Answered by Anonymous
1

❥GIVEN

 \sf  \cos x =  \frac{ - 1}{2}  \: and \:  \pi &lt; x &lt;  \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 &lt; x &lt;  \frac{3 \pi}{2}

so,

 \sf180 \degree &lt; x &lt;  \frac{3 \times 180 \degree}{2}

 \sf180 \degree &lt; x &lt;  \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  =  &gt;  \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  =  &gt; 4( \sqrt{3}  {)}^{2}  - 3( \frac{ - 2}{ \sqrt{3} }  {)}^{2}  \\  \sf  =  &gt; 4 \times 3 - 3 \times  \frac{4}{3}  \\  \sf  =  &gt; 12 - 4 = 8

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

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