How do you find the exact value of the trigonometric function given that $sin u = -sqrt(5) / 6$ and $cos v = sqrt(2) / 4$ with both u and v in quadrant IV. find sec(u+v) and Tan (u-v)?

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Answer 1 · 2017-09-02T16:34:08

$sec(u+v)=24/(sqrt62-sqrt60)$ and $tan(u-v)=(sqrt434-sqrt10)/(sqrt62+sqrt70)$

As $sinu=-sqrt5/6$ and $u$ is in $Q4$ , $cosu$ is positive and it is given by $cosu=sqrt(1-(-sqrt5/6)^2)=sqrt(1-5/36)=sqrt31/6$

Further, as $cosv=sqrt2/4$ and $v$ is in $Q4$ , $sinu$ is neegative and it is given by $sinv=-sqrt(1-(sqrt2/4)^2)=-sqrt(1-2/16)=-sqrt14/4$

Hence $cos(u+v)=cosucosv-sinusinv$

= $sqrt31/6xxsqrt2/4-(-sqrt5/6)xx(-sqrt14/4)=(sqrt62-sqrt60)/24$

and $sec(u+v)=24/(sqrt62-sqrt60)$

$cos(u-v)=cosucosv+sinusinv$

= $sqrt31/6xxsqrt2/4+(-sqrt5/6)xx(-sqrt14/4)=(sqrt62+sqrt70)/24$

and $sin(u-v)=sinucosv-cosusinv$

= $(-sqrt5/6)xxsqrt2/4-sqrt31/6xx(-sqrt14/4)=(-sqrt10+sqrt434)/24$

= $(sqrt434-sqrt10)/24$

Hence, $tan(u-v)=sin(u-v)/cos(u-v)=(sqrt434-sqrt10)/(sqrt62+sqrt70)$

How do you find the exact value of the trigonometric function given that sin u = -sqrt(5) / 6sin u = -sqrt(5) / 6 and cos v = sqrt(2) / 4cos v = sqrt(2) / 4 with both u and v in quadrant IV. find sec(u+v) and Tan (u-v)?