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Question: If \({\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \dfrac{\pi }{2}\) and \({\tan ^{ - 1}}x - {\tan ^{ - 1}}y =...

If cos⁡−1x+cos⁡−1y=π2{\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \dfrac{\pi }{2} and tan⁡−1x−tan⁡−1y=0{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = 0 then the value of x2+ax+y2{x^2} + ax + {y^2} is where a=12a = \dfrac{1}{{\sqrt 2 }}
(A) 00
(B) −12 - \dfrac{1}{2}
(C) 12\dfrac{1}{2}
(D) 32\dfrac{3}{2}

Explanation

Solution

Use the given equations cos⁡−1x+cos⁡−1y=π2{\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \dfrac{\pi }{2} and tan⁡−1x−tan⁡−1y=0{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = 0 to find the unknown values of x and yx{\text{ and }}y. Manipulate tan⁡−1x−tan⁡−1y=0{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = 0 to find a linear relationship between ‘x’ and ‘y’ then express one in terms of other. Now substitute the value of ‘y’ in terms of ‘x’ in the equation cos⁡−1x+cos⁡−1y=π2{\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \dfrac{\pi }{2}. Solve it for only the unknown. Now substitute the values of ‘x’ and ‘y’ in the expression x2+ax+y2{x^2} + ax + {y^2}.

Complete step by step solution:
Here in this problem, we are given two equations cos⁡−1x+cos⁡−1y=π2{\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \dfrac{\pi }{2} and tan⁡−1x−tan⁡−1y=0{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = 0 . Using these two equations we need to find the value for expression x2+ax+y2{x^2} + ax + {y^2} and value of ‘a’ is given as a=12a = \dfrac{1}{{\sqrt 2 }}.
In mathematics, the inverse trigonometric functions are the inverse functions of the trigonometric functions. Specifically, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are used to obtain an angle from any of the angle's trigonometric ratios. Here we are given arccosine and arctangent functions and we can use both equations to figure out the values of ‘x’ and ‘y’.
Let’s use the given equation to find the values of x and yx{\text{ and }}y and then substitute them to evaluate the required expression:
⇒tan⁡−1x−tan⁡−1y=0⇒tan⁡−1x=tan⁡−1y⇒x=y\Rightarrow {\tan ^{ - 1}}x - {\tan ^{ - 1}}y = 0 \Rightarrow {\tan ^{ - 1}}x = {\tan ^{ - 1}}y \Rightarrow x = y ………(i)
Now using the relation (i) in the second equation:
⇒cos⁡−1x+cos⁡−1y=cos⁡−1x+cos⁡−1x=2cos⁡−1x=π2⇒cos⁡−1x=π4\Rightarrow {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = {\cos ^{ - 1}}x + {\cos ^{ - 1}}x = 2{\cos ^{ - 1}}x = \dfrac{\pi }{2} \Rightarrow {\cos ^{ - 1}}x = \dfrac{\pi }{4}
As we know cos⁡π4=12⇒cos⁡−112=π4\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} \Rightarrow {\cos ^{ - 1}}\dfrac{1}{{\sqrt 2 }} = \dfrac{\pi }{4}
Therefore, we can say: cos⁡−1x=π4⇒x=12{\cos ^{ - 1}}x = \dfrac{\pi }{4} \Rightarrow x = \dfrac{1}{{\sqrt 2 }}
Thus, we got the values: x=y=12x = y = \dfrac{1}{{\sqrt 2 }}
So, the required value of the expression x2+ax+y2{x^2} + ax + {y^2} will be:
⇒x2+x2+y2=(12)2+12×12+(12)2=12+12+12=32\Rightarrow {x^2} + \dfrac{x}{{\sqrt 2 }} + {y^2} = {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }} + {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} = \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} = \dfrac{3}{2}

Note:
In questions like this, be careful with the range and domain of the functions while solving for two unknowns involving two different equations. Follow a step by step procedure to avoid any complications. Carefully substitute the value in the given expression. Inverse trigonometric functions are simply defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant functions. They are also termed as arcus functions, anti-trigonometric functions or cyclometric functions.