5(cos^2)60 + 4(sec^2)45 - (tan^2)45 = x Find x @Michele_Laina

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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is 40 degrees the angle of the second term?
45*
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? How to solve? ?
we have to keep in mind these values: \[\Large \begin{gathered} \cos 60 = \frac{1}{2} \Rightarrow {\left( {\cos 60} \right)^2} = \frac{1}{4} \hfill \\ \hfill \\ \sec 45 = \frac{1}{{\cos 45}} = \frac{1}{{\sqrt 2 /2}} = \frac{2}{{\sqrt 2 }} = \sqrt 2 \hfill \\ {\left( {\sec 45} \right)^2} = 2 \hfill \\ \hfill \\ \tan 45 = 1 \Rightarrow {\left( {\tan 45} \right)^2} = 1 \hfill \\ \end{gathered} \]
Ok!
please substitute them into your equation
OMG!!! ISY SEC 30 NETHER 40 NOR 45 :(
ok! I will rewrite the right value
OK!
here are the right values: \[\Large \begin{gathered} \cos 60 = \frac{1}{2} \Rightarrow {\left( {\cos 60} \right)^2} = \frac{1}{4} \hfill \\ \hfill \\ \sec 30 = \frac{1}{{\cos 30}} = \frac{1}{{\sqrt 3 /2}} = \frac{2}{{\sqrt 3 }} \hfill \\ {\left( {\sec 30} \right)^2} = \frac{4}{3} \hfill \\ \hfill \\ \tan 45 = 1 \Rightarrow {\left( {\tan 45} \right)^2} = 1 \hfill \\ \end{gathered} \]
so, after a simple substitution, we get: \[\Large \left( {5 \times \frac{1}{4}} \right) + \left( {4 \times \frac{4}{3}} \right) - 1 = x\]
please simplify

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