Question

Simplify the given inverse trigonometric function ${{\tan }^{-1}}\left( \dfrac{a\cos x -b\sin x}{b\cos x+a\sin x} \right)$, if $\dfrac{a}{b}\tan x >-1$.

Answer 1

Hint: At first divide the expression in the ${{\tan }^{-1}}\left( \dfrac{a\cos x- b\sin x}{b\cos x+a\sin x} \right)$ by $b\cos x$ in both numerator and denominator. Then use the identity ${{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)={{\tan }^{-1}}x-{{\tan }^{-1}}y$ and finally use the identity ${{\tan }^{-1}}\left( \tan x \right)$ = x. x and y should be greater than 0. And x lies between $-\dfrac{\pi }{2}\,\,to\,\,\dfrac{\pi }{2}$ for ${{\tan }^{-1}}x-{{\tan }^{-1}}y$

Complete step-by-step solution:
In the question we are given the expression ${{\tan }^{-1}}\left( \dfrac{a\cos x- b\sin x}{b\cos x+a\sin x} \right)$, if $\dfrac{a}{b}\tan x> -1$ and we have to simplify it.
Before proceeding let us know what are inverse trigonometric functions, trigonometric functions are functions that are inverse functions of trigonometric functions. Specifically, they are inverses of sine, cosine, tangent, cotangent, secant, cosecant functions, and are used to obtain an angle from any of the angles is trigonometric ratios.
There are certain notations which are used. Some of the most common notation is using $\arcsin \left( x \right)$, $\arccos \left( x \right)$,$\arctan \left( x \right)$ instead of ${{\sin }^{-1}}\left( x \right)$, ${{\cos }^{-1}}\left( x \right)$ and ${{\tan }^{-1}}\left( x \right)$. When measuring in radians, an angle $\theta $ radians will correspond to an arc whose length $r\theta $, where r is radius of circle. Thus, in the unit circle, “the arc whose cosine is x” is the same as “the angle whose cosine is x”, because the length of the arc of a circle in radii is the same as the measurement of angle in radius.
So, we are given expression,
${{\tan }^{-1}}\left( \dfrac{a\cos x - b\sin x}{b\cos x+a\sin x} \right)$
At first, we will analyze the expression $\dfrac{a\cos x - b\sin x}{b\cos x+a\sin x}$ and try to write in the form of tan.
We know the identity that,
$\tan \left( x- y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}$
Now to convert the denominator of $\dfrac{a\cos x -b\sin x}{b\cos x+a\sin x}$ in form of $1+\tan x\tan y$ we have to divide the numerator and denominator by $b\cos x$ so, we get the term or expression as,
$\dfrac{\dfrac{a\cos x- b\sin x}{b\cos x}}{\dfrac{b\cos x+a\sin x}{b\cos x}}$ $\Rightarrow \dfrac{\dfrac{a}{b}-\tan x}{1+\dfrac{a}{b}\tan x}$
So, the term is transformed to ${{\tan }^{-1}}\left( \dfrac{\dfrac{a}{b}-\tan x}{1+\dfrac{a}{b}\tan x} \right)$
Now, as we know the identity that ${{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)$ is equal to ${{\tan }^{-1}}x-{{\tan }^{-1}}y$ so instead of x and y we can substitute $\dfrac{a}{b},\tan x$ respectively.
So, we can write it as,
${{\tan }^{-1}}\left( \dfrac{\dfrac{a}{b}-\tan x}{1+\dfrac{a}{b}\tan x} \right)={{\tan }^{-1}}\left( \dfrac{a}{b} \right)-{{\tan }^{-1}}\left( \tan x \right)$
We know that ${{\tan }^{-1}}\left( \tan x \right)$ is x so, the find value is ${{\tan }^{-1}}\left( \dfrac{a}{b} \right)-x$.
So, on simplification we get the result as ${{\tan }^{-1}}\left( \dfrac{a}{b} \right)-x$.

Note: While solving the expression related to inverse trigonometric functions always try to get the value of expression just opposite just like in the question ${{\tan }^{-1}}$ expression was given so, try to convert the expression in tan ratios so to cancel out and get the answer.
While solving these types of questions keep in mind that we need to reduce the given equation. We need to divide the given equation by cos x and then we will convert it into tan x so that we will convert it into x tan inverse x will get canceled with the tan x.