Solveeit Logo
Login

Question

Question: How do you write an algebraic expression that is equivalent to \[\sec \left( {{\sin }^{-1}}\left( x-...

How do you write an algebraic expression that is equivalent to sec⁡(sin⁡−1(x−1))\sec \left( {{\sin }^{-1}}\left( x-1 \right) \right) ?

Explanation

Solution

These types of problems are of the topic Trigonometry and of sub topic equations and general values. We can solve this problem very quickly once we understand the underlying concepts clearly. In this problem we first of all need to convert either the one inside the bracket or the one outside the bracket into one single form. The problem would be easy if we convert the sine inverse term to secant term, then the transformation from the trigonometric term to the algebraic form would be done in no time.

Complete step by step answer:
Now we start off with the solution to the given problem by considering an imaginary right angled triangle such that the length of the perpendicular here is ‘ x−1x-1 ’ and the length of the hypotenuse is ‘ 11 ’. We assume the angle that the hypotenuse makes with the base is theta. From this information we try to find the value of the secant of the angle theta. The other side of the triangle becomes equal to 2x−x2\sqrt{2x-{{x}^{2}}} . The secant of the angle theta is therefore,
⇒sec⁡θ=12x−x2\Rightarrow \sec \theta =\dfrac{1}{\sqrt{2x-{{x}^{2}}}}
We transform the given equation as,
⇒sec⁡(sin⁡−1θ)\Rightarrow \sec \left( {{\sin }^{-1}}\theta \right)
Transforming this to secant we get,
⇒sec⁡(sec⁡−1(12x−x2))\Rightarrow \sec \left( {{\sec }^{-1}}\left( \dfrac{1}{\sqrt{2x-{{x}^{2}}}} \right) \right)
Now we know that sec⁡(sec⁡−1θ)=θ\sec \left( {{\sec }^{-1}}\theta \right)=\theta , thus we can write,
⇒sec⁡(sec⁡−1(12x−x2))=12x−x2\Rightarrow \sec \left( {{\sec }^{-1}}\left( \dfrac{1}{\sqrt{2x-{{x}^{2}}}} \right) \right)=\dfrac{1}{\sqrt{2x-{{x}^{2}}}}

Thus, our answer to the problem is 12x−x2\dfrac{1}{\sqrt{2x-{{x}^{2}}}}.

Note: We need to have a clear-cut idea regarding trigonometric equations and general values to solve these types of problems. We must also assume a right-angled triangle, doing which solving the sum becomes a bit easy. We try to transform both the inner and the outer part into one common term which would be much efficient in solving it.