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Question: Solve: \[1 + {\cot ^2}\left( {{{\sin }^{ - 1}}x} \right) = \] A. \(\dfrac{1}{{2x}}\) B. \({x^2}\...

Solve: 1+cot⁡2(sin⁡−1x)=1 + {\cot ^2}\left( {{{\sin }^{ - 1}}x} \right) =
A. 12x\dfrac{1}{{2x}}
B. x2{x^2}
C. 1x2\dfrac{1}{{{x^2}}}
D. 2x2\dfrac{2}{{{x^2}}}

Explanation

Solution

First, we will suppose sin⁡−1x{\sin ^{ - 1}}x to be equal to θ\theta . Then, xx will be equal to sin⁡θ\sin \theta . Taking θ\theta as angle of sine we will find values of sides of the triangle using the trigonometric property of triangle. Then, we will put them in the formula of cot⁡θ\cot \theta .

Complete step by step answer:
Now, let sin⁡−1=θ{\sin ^{ - 1}} = \theta
So, sin⁡θ=x\sin \theta = x
Using the trigonometric property of a triangle.
sin⁡θ=perpendicularhypotenuse\sin \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}}
And sin⁡θ=x\sin \theta = x. So, the perpendicular of the triangle is xx and the hypotenuse of the triangle is 1. By using Pythagoras theorem, we will find the value of the base of the triangle.
P2+B2=H2{P^2} + {B^2} = {H^2}

Now, we will put values of P and H in the above equation. P is equal to x and H is equal to 1.
x2+B2=12{x^2} + {B^2} = {1^2}
⇒B2=1−x2\Rightarrow {B^2} = 1 - {x^2}
⇒B=1−x2\Rightarrow B = \sqrt {1 - {x^2}}
The value of base is 1−x2\sqrt {1 - {x^2}} .Now, we will find the value of cot⁡θ\cot \theta . Taking θ\theta as angle cot we are equal to base / perpendicular.
cot⁡θ=BP\cot \theta = \dfrac{B}{P}
Now, we will put values of B and P in the above equation. B is equal to 1−x2\sqrt {1 - {x^2}} and p is equal to x.
cot⁡θ=1−x2x\cot \theta = \dfrac{{\sqrt {1 - {x^2}} }}{x}
In question there is cot⁡2{\cot ^2}. So, we will square both the sides.
cot⁡2θ=1−x2x2{\cot ^2}\theta = \dfrac{{1 - {x^2}}}{{{x^2}}}

Now, we have all the required values. So, we will put them in the equation given in question.
⇒1+cot⁡2(sin⁡−1x)\Rightarrow 1 + {\cot ^2}\left( {{{\sin }^{ - 1}}x} \right)
Earlier we let sin⁡−1x=θ{\sin ^{ - 1}}x = \theta . So, we will replace it in the above equation.
⇒1+cot⁡2θ\Rightarrow 1 + {\cot ^2}\theta
Now we will put the value of cot⁡2θ{\cot ^2}\theta in the above equation.
⇒1+1−x2x2\Rightarrow 1 + \dfrac{{1 - {x^2}}}{{{x^2}}}
⇒x2+1−x2x2\Rightarrow \dfrac{{{x^2} + 1 - {x^2}}}{{{x^2}}}
∴1x2\therefore \dfrac{1}{{{x^2}}}
The value of 1+cot⁡2(sin⁡−1x)1 + {\cot ^2}\left( {{{\sin }^{ - 1}}x} \right) is 1x2\dfrac{1}{{{x^2}}}.

Therefore, option C is the correct answer.

Note: While finding the square root of any squared term, do not neglect the negative terms. Both positive and negative values are required. Students should learn all the basic trigonometric identities and functions for solving these questions. Pythagoras theorem establishes a relation between the sides of a right angled triangle and helps us to find cotangent of an angle whose sine is given to us.