1. given \\( \\tan \\theta=-\\frac{9}{4} \\), where \\( 270^{\\circ} \\leq \\theta \\leq 360^{\\circ}…

1. given \\( \\tan \\theta=-\\frac{9}{4} \\), where \\( 270^{\\circ} \\leq \\theta \\leq 360^{\\circ} \\),\na) state the other five trigonometric ratios as fractions. 5 marks\n\\( 270^{\\circ} \\leq \\theta \\leq 360^{\\circ} \\)\n\\( \\sin \\theta<0, \\cos \\theta>0, \\tan \\theta<0, \\csc \\theta<0, \\sec \\theta>0 \\),\n\\( \\cot \\theta<0. \\quad \\tan \\theta=-\\frac{9}{4}=\\frac{y}{x} \\)\n\\( \\tan \\theta=\\frac{y}{x}, \\quad x = 4, \\quad y=-9 \\)\n\\( r = \\sqrt { x ^ { 2 } + y ^ { 2 } } = r = \\sqrt { 4 ^ { 2 } + ( - 9 ) ^ { 2 } } = \\sqrt { 16 + 81 } = \\sqrt { 97 } \\).\n\\( \\theta = \\frac { y } { r } = y=-9, r = \\sqrt { 97 } \\).\n\\( \\sin \\theta = \\frac { - 9 } { \\sqrt { 97 } } = - \\frac { 9 \\sqrt { 97 } } { 97 } \\);\nb) determine the value of \\( \\theta \\) to the nearest degree. 1 mark

1. given \\( \\tan \\theta=-\\frac{9}{4} \\), where \\( 270^{\\circ} \\leq \\theta \\leq 360^{\\circ} \\),\na) state the other five trigonometric ratios as fractions. 5 marks\n\\( 270^{\\circ} \\leq \\theta \\leq 360^{\\circ} \\)\n\\( \\sin \\theta<0, \\cos \\theta>0, \\tan \\theta<0, \\csc \\theta<0, \\sec \\theta>0 \\),\n\\( \\cot \\theta<0. \\quad \\tan \\theta=-\\frac{9}{4}=\\frac{y}{x} \\)\n\\( \\tan \\theta=\\frac{y}{x}, \\quad x = 4, \\quad y=-9 \\)\n\\( r = \\sqrt { x ^ { 2 } + y ^ { 2 } } = r = \\sqrt { 4 ^ { 2 } + ( - 9 ) ^ { 2 } } = \\sqrt { 16 + 81 } = \\sqrt { 97 } \\).\n\\( \\theta = \\frac { y } { r } = y=-9, r = \\sqrt { 97 } \\).\n\\( \\sin \\theta = \\frac { - 9 } { \\sqrt { 97 } } = - \\frac { 9 \\sqrt { 97 } } { 97 } \\);\nb) determine the value of \\( \\theta \\) to the nearest degree. 1 mark

Answer

Explanation:

Step1: Recall the formula for ( \cot\theta )

Since ( \cot\theta=\frac{1}{\tan\theta} ), and ( \tan\theta =-\frac{9}{4} ), then ( \cot\theta=-\frac{4}{9} )

Step2: Use the Pythagorean identity ( \sec^{2}\theta = 1+\tan^{2}\theta )

[ \begin{align*} \sec^{2}\theta&=1+\left(-\frac{9}{4}\right)^{2}\ &=1 + \frac{81}{16}\ &=\frac{16 + 81}{16}\ &=\frac{97}{16} \end{align*} ] Since ( 270^{\circ}\leq\theta\leq360^{\circ} ), ( \cos\theta>0 ) and ( \sec\theta>0 ). So ( \sec\theta=\frac{\sqrt{97}}{4} )

Step3: Find ( \cos\theta )

Since ( \cos\theta=\frac{1}{\sec\theta} ), then ( \cos\theta=\frac{4}{\sqrt{97}}=\frac{4\sqrt{97}}{97} )

Step4: Use the identity ( \sin^{2}\theta+\cos^{2}\theta = 1 )

[ \begin{align*} \sin^{2}\theta&=1-\cos^{2}\theta\ &=1-\frac{16}{97}\ &=\frac{97-16}{97}\ &=\frac{81}{97} \end{align*} ] Since ( 270^{\circ}\leq\theta\leq360^{\circ} ), ( \sin\theta<0 ). So ( \sin\theta=-\frac{9}{\sqrt{97}}=-\frac{9\sqrt{97}}{97} )

Step5: Find ( \csc\theta )

Since ( \csc\theta=\frac{1}{\sin\theta} ), then ( \csc\theta=-\frac{\sqrt{97}}{9} )

Answer:

(\sin\theta =-\frac{9\sqrt{97}}{97}), (\cos\theta=\frac{4\sqrt{97}}{97}), (\csc\theta =-\frac{\sqrt{97}}{9}), (\sec\theta=\frac{\sqrt{97}}{4}), (\cot\theta=-\frac{4}{9})