Question

if tan alpha=3 and sin beta=1/3. Then find cos (alpha-beta). Where alpha nad beta lies in the third and second quadrant.

Answer 1

Answer:

$\frac{4 \sqrt{5} - 3 \sqrt{10} }{30}$

Step-by-step explanation:

$\tan \alpha = 3 \\ ⇒ { \tan }^{2} \alpha + 1 = { \sec }^{2} \alpha \\ ⇒9 + 1 = {sec}^{2} \alpha \\ ⇒ \sec \alpha = - \sqrt{10} \: as \: secant \: is \: negative \: in \: 3rd \: quad. \\ ⇒cos \alpha = \frac{ - 1}{ \sqrt{10} } \\ sin \alpha = - \sqrt{1 - \frac{1}{10} } \\ ⇒sin \: \alpha = \frac{ - 3}{ \sqrt{10} } \\ sin \: \beta = \frac{1}{3} \\ cos \: \beta = - \sqrt{1 - \frac{1}{9} } = \frac{ - 2 \sqrt{2} }{3} \\ cos( \alpha - \beta ) =cos \: \alpha \: cos \: \beta + sin \: \alpha \: sin \: \beta \\ ⇒cos( \alpha - \beta ) = \frac{ - 1}{ \sqrt{10} } \times \frac{ - 2 \sqrt{2} }{3} + \frac{ - 3}{ \sqrt{10} } \times \frac{1}{3} \\ ⇒cos( \alpha - \beta ) = \frac{2 \sqrt{2} - 3 }{3 \sqrt{10} }$

AFTER RATIONALIZATION,

$cos( \alpha - \beta ) = \frac{4 \sqrt{5} - 3 \sqrt{10} }{30}$

Answer 2

Answer:

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