Step-by-step explanation:
1.
[tex] \tan \beta = \frac{1}{ \cot \beta } = - \frac{3}{2 \sqrt{10} } = - \frac{3 \sqrt{10} }{20} [/tex]
[tex] \csc \beta \tan \beta = \frac{1}{ \cos \beta } = \sec \beta [/tex]
Therefore,
[tex] \sec \beta = ( \frac{7}{3} )( - \frac{3 \sqrt{10} }{20} ) = - \frac{7 \sqrt{10} }{20} [/tex]
2.
[tex] \csc y = \frac{1}{ \sin y} = - \frac{ \sqrt{6} }{2} [/tex]
[tex] = > \sin y = - \frac{ \sqrt{6} }{3} [/tex]
Use the identity
[tex] \cos y = \sqrt{1 - \sin ^{2} y} \: \: \: \: \: \: \: \: \: \: \: \\ = \sqrt{1 - {( - \frac{ \sqrt{6} }{3}) }^{2} } = - \frac{ \sqrt{3} }{3} [/tex]
We chose the negative value of the cosine because of the condition where cot y > 0. Otherwise, choosing the positive root will yield a negative cotangent value. Now that we know the sine and cosine of y, we can now solve for the tangent:
[tex]\tan \beta = \frac{ \sin y}{ \cos y} =( - \frac{ \sqrt{6} }{3} )( - \frac{3}{ \sqrt{3} } ) = \sqrt{2} [/tex]
3. Recall that sec x = 1/cos x, therefore cos x = 5/6. Solving for sin x,
[tex] \sin x = \sqrt{1 - \cos ^{2} x} = \sqrt{ \frac{11}{6} } [/tex]
Solving for tan x:
[tex] \tan x = \frac{ \sin x}{ \cos x} = (\frac{ \sqrt{11} }{ \sqrt{6} } )( \frac{6}{5} ) = \frac{ \sqrt{66} }{5} [/tex]
