I would certainly like to recognize if there is a basic technique to fix equation looking choose this:

\$\$ an(sec^-1 4)\$\$

without making use of a calculator (you need to discover the precise value)?

How to proceed?  Imagine a right-angled triangle through one leg \$k\$ and hypotenuse \$4k\$ and angle \$ heta\$ in between them. Then \$cos heta = frack4k= frac14\$ and \$sec heta = 4\$, making \$sec^-14 = heta\$.

The opposite leg is \$sqrt(4k)^2-k^2=sqrt15k\$ and so \$ an(sec^-14) = an heta = fracsqrt15kk=sqrt15\$. Now you might need a calculator.

You are watching: Sec^-1(4) Let \$sec^-14= hetaimpliessec heta=4\$

Now, \$ an^2 heta=sec^2 heta-1\$

Finally utilizing the definition of the primary value of \$sec^-1,00\$ Thanks for contributing a solution to steustatiushistory.orgematics Stack Exchange!

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If \$2 an^2 x - 5sec x - 1 = 0\$ has 7 roots in \$< 0,fracnpi2>\$ then the biggest worth of \$n\$ is?
Solving the value of \$x\$ for the equation \$frac (sec x + an x)^2 - (sec 2x + an x)^2sin 2x - sin x = 2\$ website architecture / logo design © 2021 Stack Exreadjust Inc; user contributions licensed under cc by-sa. rev2021.9.28.40331