Question

sin⁻¹(cos(sin⁻¹x))+ cos⁻¹(sin(cos⁻¹x)) is.......,Select Proper option from the given options.
(a) 0
(b) π/4
(c) π/2
(d) 3π/4

Answer 1

Dear Student,

Answer: π/2 ( option c)

Solution:

1) convert sin⁻¹x to cos⁻¹x, so that cos cancels cos⁻¹x

2) In second half convert cos⁻¹x into sin⁻¹x ,so that sin cancels sin⁻¹x

3) apply formula sin⁻¹x+cos⁻¹x =π/2

${sin}^{ - 1} ( \cos( {sin}^{ - 1} x) + {cos}^{ - 1} ( \sin({cos}^{ - 1} x) \\ {sin}^{ - 1} ( \cos({sin}^{ - 1} x) = {sin}^{ - 1} (cos( {cos}^{ - 1} \sqrt{1 - {x}^{2} } ) \\ \\ = {sin}^{ - 1} \sqrt{1 - {x}^{2} } \\ \\ {cos}^{ - 1} ( \sin( {cos}^{ - 1} x) \\ = {cos}^{ - 1} (sin( {sin}^{ - 1} \sqrt{1 - {x}^{2} } ) \\ \\ = {cos}^{ - 1} \sqrt{1 - {x}^{2} } \\ \\ {sin}^{ - 1} ( \cos({sin}^{ - 1} x) + {cos}^{ - 1} ( \sin( {cos}^{ - 1} x) \\ \\ = {sin}^{ - 1} \sqrt{1 - {x}^{2} } + {cos}^{ - 1} \sqrt{1 - {x}^{2} }$
since sin⁻¹x+cos⁻¹x =π/2

So,
${sin}^{ - 1} \sqrt{1 - {x}^{2} } + {cos}^{ - 1} \sqrt{1 - {x}^{2} } = \frac{\pi}{2}$

So,the answer is π/2

Hope it helps you

Answer 2

we have to find the value of sin^-1(cos(sin^-1x)) + cos^-1(sin(cos^-1x)) .........(1)

Let sin^-1x = A
sinA =x => cosA = √(1 - x^2)
cos^-1(√(1 - x^2)) = A
so, sin^-1(cos(cos^-1√(1-x^2))= sin^-1(√(1-x^2).....(2)

similarly, cos^-1x = B
cosB = x => sinB = √(1 - x^2)
sin^-1√(1 - x^2) = B
so, cos^-1(sin(sin^-1√(1-x^2)) = cos^-1√(1-x^2)......(3)

put equations (2) and (3), in equation (1)
sin^-1(cos(sin^-1x)) + cos^-1(sin(cos^-1x))= sin^-1{√(1 - x^2)} + cos^-1{√(1 - x^2)}

we know, sin^-1z + cos^-1z = π/2 , where |z| ≤ 1

so, sin^-1{√(1 - x^2)} + cos^-1{√(1 - x^2)} = π/2
because √(1 - x^2) ≤ 1

hence, option(c) is correct