Question

If cosx is -1/2 then find remaining trigonometry ratios.​

Answer 1

Answer:

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Step-by-step explanation:

Used Right Angled Triangle for explaining

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Answer 2

${\underline{\boxed{\sf\bold \red{Solution}}}}$

$\sf \: cos \: X= \frac{ - 1}{2} = \tt\large\red{ \frac{adjacent \: side}{hypotenuse}} = \sf \: \frac{XY}{XZ}$

$\sf \: By \: Puthygoras \: theorem, \\ \\ \sf \: (XZ)² = (XY)² + (YZ)² \\ \sf \: (2k)² = (-1k)² + (YZ)² \\ \sf \: 4k² = 1k² + YZ² \\ \sf \: 4k² - 1k² = YZ² \\ \sf \: 3k² = YZ² \\ \sf \: YZ^{2}=3k^{2} \\ \sf \: YZ = \sqrt{3k^{2}} \\ \sf \: YZ=\sqrt{3k}$

$\rm\underline{Other \: Trigonometric \: ratios \: are :}$

$\sf \: sin X = \frac{opposite \: side}{hypotenuse} = \frac{YZ}{XZ} = \frac{\sqrt3{ \cancel{k}}}{2{ \cancel{k}}} = \frac{\sqrt3}{2}$

$\sf \: tan X = \frac{opposite \: side}{adjacent \: side}= \frac{YZ}{XY} = \frac{\sqrt3{ \cancel{k}}}{ - 1{ \cancel{k}}} = \frac{\sqrt3}{ - 1}$

$\sf \: cosec X = \frac{hypotenuse}{opposite \: side }= \frac{XZ}{YZ} =\frac{2{ \cancel{k}}}{\sqrt3{ \cancel{k}}} = \frac{2}{\sqrt3}$

$\sf \: sec X = \frac{hypotenuse}{adjacent \: side}=\frac{XZ}{XY} = \frac{2{ \cancel{k}}}{ - 1{ \cancel{k}}} = \frac{2}{ - 1}$

$\sf \: cot X = \frac{adjacent \: side}{opposite \: side}=\frac{XY}{YZ} = \frac{ - 1{ \cancel{k}}}{\sqrt3{ \cancel{k}}} = \frac{ - 1}{\sqrt3}$