If csc A= -5/3 and csc B = -13/5 with angle A and angle B in quadrant 3. What is value of cos(A-B)?

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If csc A= -5/3 and csc B = -13/5 with angle A and angle B in quadrant 3. What is value of cos(A-B)?

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Answer 1 · 2016-07-23T17:16:11

$cos (A - B) = \frac{63}{65}$ $cos(A-B)=63/65$ .

Explanation:

We start by taking note that angles $A and B$ $A and B$ are in $Q_{I I I}$ $Q_(III$ , so that, $cos A and cos B$ $cosA and cosB$ both will be $- v e$ $-ve$ .

$csc A = - \frac{5}{3} \Rightarrow sin A = - \frac{3}{5} \Rightarrow cos A = - \sqrt{1 - {sin}^{2} A} = - \sqrt{1 - \frac{9}{25}} = - \frac{\sqrt{16}}{25} = - \frac{4}{5}$ $cscA=-5/3rArrsinA=-3/5rArrcosA=-sqrt(1-sin^2A)=-sqrt(1-9/25)=-sqrt16/25=-4/5$

Similarly, $csc B = - \frac{13}{5} \Rightarrow cos B = - \frac{12}{13}$ $cscB=-13/5rArrcosB=-12/13$

Therefore, reqd. value $= cos (A - B) = cos A cos B + sin A sin B$ $=cos(A-B)=cosAcosB+sinAsinB$

$= (- \frac{4}{5}) (- \frac{12}{13}) + (- \frac{3}{5}) (- \frac{5}{13})$ $=(-4/5)(-12/13)+(-3/5)(-5/13)$

$\frac{48}{65} + \frac{15}{65} = \frac{63}{65}$ $48/65+15/65=63/65$ .

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If csc A= -5/3 and csc B = -13/5 with angle A and angle B in quadrant 3. What is value of cos(A-B)?

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Explanation:

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