Question 1186337
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If λ is an angle in the second quadrant and sec λ = - √85/6. 
Find the value for cos^4 λ-2sin^2 λ +cot^3 λ, as a decimal.
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To facilitate my writing, I will use variable x instead of "  λ ".


Since sec(x) = {{{-sqrt(85)/6}}},  we have  cos(x) = {{{1/sec(x)}}} = {{{-6/sqrt(85)}}}.


Next, sin(x) = {{{sqrt(1 - cos^2(x))}}} = {{{sqrt(1 - (-6/sart(85))^2)}}} = {{{sqrt(1 - 36/85)}}} = {{{sqrt(49/85)}}} = {{{7/sqrt(85)}}}.


      (in my derivation, I used the fact that the angle x is in the second quadrant, taking the sign "+" at the square root).


So, cos(x) = {{{-6/sqrt(85)}}};  sin(x) = {{{7/sqrt(85)}}};  cot(x) = {{{cos(x)/sin(x)}}} = {{{-6/7}}}.


Now  cos^4 (x)-2sin^2 (x) +cot^3 (x) = {{{(-6/sqrt(85))^4 - 2*(7/sqrt(85))^2 + (-6/7)^3}}} = use your calculator = 1.6033    (rounded).    <U>ANSWER</U>
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Solved.