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Question: How do you prove \[{\sin ^{ - 1}}\left( x \right) + {\cos ^{ - 1}}\left( x \right) = \dfrac{\pi }{2}...

How do you prove sin⁡−1(x)+cos⁡−1(x)=π2{\sin ^{ - 1}}\left( x \right) + {\cos ^{ - 1}}\left( x \right) = \dfrac{\pi }{2} ?

Explanation

Solution

Hint : Here in this question we have to prove the given function. The function is related to inverse trigonometry. When the trigonometry ratio is raised to the power -1 then it is an inverse trigonometry. By applying the ASTC rule and substituting the trigonometry ratio or function to some variable we are going to prove the given function

Complete step-by-step answer :
ASTC rule of trigonometry ASTC rule stands for the "all sine tangent cosine" rule. It is intended to remind us that all trigonometric ratios are positive in the first quadrant of a graph, only the sine and its cofunction cosecant are positive in the second quadrant, only the tangent and its cofunction cotangent are positive in the third quadrant, and only the cosine and its cofunction secant are positive in the fourth quadrant. One way to remember this arrangement is with a sentence “All students take coffee” or “All science teachers are crazy”.
Then always remember, when you write the trigonometric function with angle 90∘{90^ \circ } or 270∘{270^ \circ } , the function will change to its cofunction.
Consider the given equation
⇒sin⁡−1(x)+cos⁡−1(x)=π2\Rightarrow {\sin ^{ - 1}}\left( x \right) + {\cos ^{ - 1}}\left( x \right) = \dfrac{\pi }{2}
Let consider
A=sin⁡−1(x)A = {\sin ^{ - 1}}\left( x \right) and B=cos⁡−1(x)B = {\cos ^{ - 1}}\left( x \right)
x=sin⁡Ax = \sin A x=cos⁡Bx = \cos B
Therefore,
x=sin⁡A=cos⁡Bx = \sin A = \cos B
Or
sin⁡A=cos⁡B\sin A = \cos B
By using ASTC rule cos⁡B\cos B can be written as sin⁡(90−B)\sin \left( {90 - B} \right) , then
⇒sin⁡A=sin⁡(90−B)\Rightarrow \sin A = \sin \left( {90 - B} \right)
Take sin⁡−1{\sin ^{ - 1}} on both sides
⇒sin⁡−1(sin⁡A)=sin⁡−1(sin⁡(90−B))\Rightarrow {\sin ^{ - 1}}\left( {\sin A} \right) = {\sin ^{ - 1}}\left( {\sin \left( {90 - B} \right)} \right)
As we know the x.x−1=1x.{x^{ - 1}} = 1 , then
⇒A=90−B\Rightarrow A = 90 - B
Add B on both sides
⇒A+B=90−B+B\Rightarrow A + B = 90 - B + B
⇒A+B=90\Rightarrow A + B = 90
Substitute A and B value
⇒sin⁡−1(x)+cos⁡−1(x)=90\Rightarrow {\sin ^{ - 1}}\left( x \right) + {\cos ^{ - 1}}\left( x \right) = 90
Convert 90 degree to radian by multiplying π180\dfrac{\pi }{{180}} ⇒90×π180=π2c \Rightarrow 90 \times \dfrac{\pi }{{180}} = {\dfrac{\pi }{2}^c}
∴sin⁡−1(x)+cos⁡−1(x)=π2\therefore {\sin ^{ - 1}}\left( x \right) + {\cos ^{ - 1}}\left( x \right) = \dfrac{\pi }{2}
Hence proved.
So, the correct answer is “sin⁡−1(x)+cos⁡−1(x)=π2{\sin ^{ - 1}}\left( x \right) + {\cos ^{ - 1}}\left( x \right) = \dfrac{\pi }{2}”.

Note : In trigonometry and inverse trigonometry we have a table for trigonometry ratios for standard angles. By using the table, we can determine the values. The inverse for the trigonometry ratio is represented by arc or trigonometry ratio is raised by -1. Hence we can solve these types of questions by knowing the table of trigonometry ratios for standard angles and the ASTC rule