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Question: If an inverse trigonometric equation is given as \({{\cos }^{-1}}x-{{\sin }^{-1}}x=0\) , then \(x\) ...

If an inverse trigonometric equation is given as cos⁡−1x−sin⁡−1x=0{{\cos }^{-1}}x-{{\sin }^{-1}}x=0 , then xx is equal to.
(a) ±12\pm \dfrac{1}{\sqrt{2}}
(b) 11
(c) ±13\pm \dfrac{1}{\sqrt{3}}
(d) 12\dfrac{1}{\sqrt{2}}

Explanation

Solution

Hint: For solving this question first we will prove an important formula of inverse trigonometric functions i.e., sin⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}. After that, we will find the angle sin⁡−1x{{\sin }^{-1}}x and then solve further to find the exact value of xx.

Complete step-by-step solution -
Given:
It is given that, cos⁡−1x−sin⁡−1x=0{{\cos }^{-1}}x-{{\sin }^{-1}}x=0 and we have to find the value of xx.
Now, let sin⁡θ=x\sin \theta =x. Then,
sin⁡θ=x ⇒θ=sin⁡−1x..................(1) \begin{aligned} & \sin \theta =x \\\ & \Rightarrow \theta ={{\sin }^{-1}}x..................\left( 1 \right) \\\ \end{aligned}
Now, as we know that range of function θ=sin⁡−1x\theta ={{\sin }^{-1}}x will be from [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] . Then,
−π2≤θ≤π2...................(2)-\dfrac{\pi }{2}\le \theta \le \dfrac{\pi }{2}...................\left( 2 \right)
Now, we know that sin⁡θ=cos⁡(π2−θ)\sin \theta =\cos \left( \dfrac{\pi }{2}-\theta \right) always. Then,
sin⁡θ=x ⇒cos⁡(π2−θ)=x \begin{aligned} & \sin \theta =x \\\ & \Rightarrow \cos \left( \dfrac{\pi }{2}-\theta \right)=x \\\ \end{aligned}
Now, let π2−θ=α\dfrac{\pi }{2}-\theta =\alpha so, we can write cos⁡(π2−θ)=cos⁡α\cos \left( \dfrac{\pi }{2}-\theta \right)=\cos \alpha in the above equation. Then,
cos⁡(π2−θ)=x ⇒cos⁡α=x ⇒α=cos⁡−1x.................(3) \begin{aligned} & \cos \left( \dfrac{\pi }{2}-\theta \right)=x \\\ & \Rightarrow \cos \alpha =x \\\ & \Rightarrow \alpha ={{\cos }^{-1}}x.................\left( 3 \right) \\\ \end{aligned}
Now, as we know that range of function α=cos⁡−1x\alpha ={{\cos }^{-1}}x will be from [0,π]\left[ 0,\pi \right] . Then,
0≤α≤π0\le \alpha \le \pi
Now, as per our assumption π2−θ=α\dfrac{\pi }{2}-\theta =\alpha so, we can write α=π2−θ\alpha =\dfrac{\pi }{2}-\theta in the above inequality. Then,
0≤α≤π ⇒0≤π2−θ≤π \begin{aligned} & 0\le \alpha \le \pi \\\ & \Rightarrow 0\le \dfrac{\pi }{2}-\theta \le \pi \\\ \end{aligned}
Now, subtract π2\dfrac{\pi }{2} in the above inequality. Then,
0≤π2−θ≤π ⇒−π2≤π2−θ−π2≤π−π2 ⇒−π2≤−θ≤π2 \begin{aligned} & 0\le \dfrac{\pi }{2}-\theta \le \pi \\\ & \Rightarrow -\dfrac{\pi }{2}\le \dfrac{\pi }{2}-\theta -\dfrac{\pi }{2}\le \pi -\dfrac{\pi }{2} \\\ & \Rightarrow -\dfrac{\pi }{2}\le -\theta \le \dfrac{\pi }{2} \\\ \end{aligned}
Now, multiply by -1 in the above inequality. Then,
−π2≤−θ≤π2 ⇒π2≥θ≥−π2 \begin{aligned} & -\dfrac{\pi }{2}\le -\theta \le \dfrac{\pi }{2} \\\ & \Rightarrow \dfrac{\pi }{2}\ge \theta \ge -\dfrac{\pi }{2} \\\ \end{aligned}
Now, we can write the above inequality as −π2≤−θ≤π2-\dfrac{\pi }{2}\le -\theta \le \dfrac{\pi }{2} and as this result matches with the equation (2) so, we conclude that we can write α=cos⁡−1x\alpha ={{\cos }^{-1}}x for π2−θ=α\dfrac{\pi }{2}-\theta =\alpha always. Then,
α=cos⁡−1x ⇒π2−θ=cos⁡−1x \begin{aligned} & \alpha ={{\cos }^{-1}}x \\\ & \Rightarrow \dfrac{\pi }{2}-\theta ={{\cos }^{-1}}x \\\ \end{aligned}
Now, put the value of θ=sin⁡−1x\theta ={{\sin }^{-1}}x from equation (1) in the above equation. Then,
π2−θ=cos⁡−1x ⇒π2−sin⁡−1x=cos⁡−1x ⇒sin⁡−1x+cos⁡−1x=π2 \begin{aligned} & \dfrac{\pi }{2}-\theta ={{\cos }^{-1}}x \\\ & \Rightarrow \dfrac{\pi }{2}-{{\sin }^{-1}}x={{\cos }^{-1}}x \\\ & \Rightarrow {{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} \\\ \end{aligned}
Now, from the above result, we conclude that sin⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} is always true. Then,
sin⁡−1x+cos⁡−1x=π2 ⇒cos⁡−1x=π2−sin⁡−1x \begin{aligned} & {{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} \\\ & \Rightarrow {{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x \\\ \end{aligned}
Now, subtract sin⁡−1x{{\sin }^{-1}}x from the above equation. Then,
cos⁡−1x=π2−sin⁡−1x ⇒cos⁡−1x−sin⁡−1x=π2−sin⁡−1x−sin⁡−1x ⇒cos⁡−1x−sin⁡−1x=π2−2sin⁡−1x \begin{aligned} & {{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x \\\ & \Rightarrow {{\cos }^{-1}}x-{{\sin }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x-{{\sin }^{-1}}x \\\ & \Rightarrow {{\cos }^{-1}}x-{{\sin }^{-1}}x=\dfrac{\pi }{2}-2{{\sin }^{-1}}x \\\ \end{aligned}
Now, it is given that cos⁡−1x−sin⁡−1x=0{{\cos }^{-1}}x-{{\sin }^{-1}}x=0 . Then,
cos⁡−1x−sin⁡−1x=π2−2sin⁡−1x ⇒0=π2−2sin⁡−1x ⇒2sin⁡−1x=π2 ⇒sin⁡−1x=π4 \begin{aligned} & {{\cos }^{-1}}x-{{\sin }^{-1}}x=\dfrac{\pi }{2}-2{{\sin }^{-1}}x \\\ & \Rightarrow 0=\dfrac{\pi }{2}-2{{\sin }^{-1}}x \\\ & \Rightarrow 2{{\sin }^{-1}}x=\dfrac{\pi }{2} \\\ & \Rightarrow {{\sin }^{-1}}x=\dfrac{\pi }{4} \\\ \end{aligned}
Now, apply sine operator in the above equation and then, write sin⁡(sin⁡−1x)=x\sin \left( {{\sin }^{-1}}x \right)=x and sin⁡π4=12\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} . Then,
sin⁡−1x=π4 ⇒sin⁡(sin⁡−1x)=sin⁡π4 ⇒x=12 \begin{aligned} & {{\sin }^{-1}}x=\dfrac{\pi }{4} \\\ & \Rightarrow \sin \left( {{\sin }^{-1}}x \right)=\sin \dfrac{\pi }{4} \\\ & \Rightarrow x=\dfrac{1}{\sqrt{2}} \\\ \end{aligned}
Now, from the above result, we conclude that if cos⁡−1x−sin⁡−1x=0{{\cos }^{-1}}x-{{\sin }^{-1}}x=0 , then the value of x=12x=\dfrac{1}{\sqrt{2}} .
Hence, option (d) will be the correct option.

Note: Here, the student should first understand and then proceed in the right direction to get the correct answer quickly. Moreover, for objective problems, we should remember the result sin⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} and apply it directly without any mistake to get the correct answer quickly without doing lengthy calculations.