Express sin - 1 x in terms of (i) cos - 1 1 - x 2 (ii) tan - 1 x 1 - x 2 (iii) cot - 1 1 - x 2 x

To express sin - 1 x in terms of another inverse trigonometric expression. Let consider sin - 1 x = θ Domain = x ∈ - 1 , 1 ⇒ sin θ = x ⇒ sin θ = x 1 . . . ( i ) Now, we can draw a right triangle by using the equation ( i ) from the triangle, we have cos θ = 1 - x 2 ⇒ θ = cos - 1 1 - x 2 . . . ( ii ) tan θ = x 1 - x 2 ⇒ θ = tan - 1 x 1 - x 2 . . . ( iii ) cot θ = 1 - x 2 x ⇒ θ = cot - 1 1 + x 2 x . . . ( iv ) sin θ , we know sin - 1 x = θ , where x ∈ [ - 1 , 1 ] (i) ⇒ cos - 1 1 - x 2 , 0 ≤ x ≤ 1 - cos - 1 1 - x 2 , - 1 ≤ x < 0 (ii) ⇒ tan - 1 x 1 - x 2 , x ∈ ( - 1 , 1 ) for all (iii) ⇒ cot - 1 1 - x 2 x 0 < x ≤ 1 cot - 1 1 - x 2 x - 1 ≤ x < 0

sin - 1 x 2 4 + y 2 9 + cos - 1 x 2 2 + y 3 2 - 2 equals to

We know, in sin - 1 t . . . i t ∈ [ - 1 , 1 ] But, here given sin - 1 x 2 4 + y 2 9 . . . ii Comparing ( i ) & ( ii ) , we get - 1 ≤ x 2 4 + y 2 9 ≤ 1 The above equation shows an area inside the ellipse x 2 4 + y 2 9 and equation of a tangent to the ellipse is y = m x ± a 2 m 2 + b 2 . . . iii Similarly in cos - 1 t . . . ( iv ) t ∈ [ - 1 , 1 ] But here given, cos - 1 x 2 2 + y 3 2 - 2 . . . v Comparing iv & v , we get - 1 ≤ x 2 2 + y 3 2 - 2 ≤ 1 ⇒ x 2 2 + y 3 2 = 1 & x 2 2 + y 3 2 = 3 . . . vi By comparing ( iii ) & ( vi ) , we draw the above two lines. Equation of a tangent to the ellipse is y = m x ± a 2 m 2 + b 2 . . . iii Therefore, intersection point P which comes in both region. Since, point P lies on the ellipse x 2 4 + y 2 9 = 1 P = ( 2 cos θ , 2 sin θ ) P satisfies the ellipse tangent equation ⇒ x x 1 4 + y y 1 9 = 1 ⇒ x cos θ 2 + y sin θ 3 = 1 But, we get the tangent equation as x 2 2 + y 3 2 = 1 (from the equation ( vi ) ) By comparing the coefficient of these equations, we get cos θ = 1 2 & sin θ = 1 2 P = 2 , 3 3 = ( x , y ) It satisfies the given question. Now, putting point P in that equation we get P = 2 , 3 3 = ( x , y ) sin - 1 x 2 4 + y 2 9 + cos - 1 x 2 2 + y 3 2 - 2 = sin - 1 1 + cos - 1 1 2 + 1 2 - 2 = sin - 1 1 + cos - 1 - 1 = π 2 + π = 3 π 2

If α = 2 tan - 1 1 + x 1 - x & β = sin - 1 1 - x 2 1 + x 2 for 0 < x < 1 , then prove that α + β = π . What the value of α + β will be if x > 1 ?

We have α = 2 tan - 1 1 + x 1 - x & β = sin - 1 1 - x 2 1 + x 2 for 0 < x < 1 Let x = tan θ , where 0 < θ < π 4 Then, α = 2 tan - 1 tan π 4 + tan θ 1 - tan π 4 tan θ = 2 tan - 1 . tan θ + π 4 As we know, tan - 1 tan ( A ) = A ⇒ α = 2 θ + π 4 = 2 θ + π 2 . . . ( i ) Similarly, β = sin - 1 1 - tan 2 θ 1 + tan 2 θ = sin - 1 ( cos 2 θ ) = sin - 1 sin π 2 - 2 θ As we know, sin - 1 sin ( A ) = A ⇒ β = π 2 - 2 θ . . . ( ii ) From ( i ) & ( ii ) , we get α + β = 2 θ + π 2 + π 2 - 2 θ α + β = π Hence proved. Now, to find α + β for x > 1 Again, let x = tan θ , where θ = π 4 Then. α = 2 tan - 1 tan θ + π 4 - π = 2 θ + π 4 - π ⇒ α = 2 θ - 3 π 2 Similarly, β = sin - 1 sin π 2 - 2 θ ⇒ β = π 2 - 2 θ ( ∵ No change in β ) α + β = - π

Find the solution of sin - 1 x 1 + x - sin - 1 x - 1 x + 1 = sin - 1 1 1 + x

sin - 1 x 1 + x - sin - 1 x - 1 x + 1 = sin - 1 1 1 + x For domain of x , ∴ x 1 + x ≥ 0 and 1 + x > 0 ; square root is never negative. ∴ x ≥ 0 or x < - 1 and x > - 1 On taking intersection of above cases, we get, ∴ x ≥ 0 (values of x for which the given equation is defined) We know, sin - 1 x - sin - 1 y = sin - 1 x 1 - y 2 - y 1 - x 2 Applying above identity, Now, LHS = sin - 1 x 1 + x 1 - x - 1 x + 1 2 - x - 1 x + 1 1 - x 1 + x = sin - 1 x 1 + x × ( x + 1 ) 2 - ( x - 1 ) 2 ( x + 1 ) 2 - x - 1 x + 1 1 1 + x = sin - 1 2 x ( x + 1 ) 3 / 2 - ( x - 1 ) ( x + 1 ) 3 / 2 = sin - 1 x + 1 ( x + 1 ) 3 / 2 = sin - 1 1 1 + x = RHS Hence, all values of x which are in the domain will be the solution of the given equation, since we have shown that LHS = RHS . ∴ Final solution : x ≥ 0

Determine the integral values of ' k ' for which the system tan - 1 x 2 + cos - 1 y 2 = π 2 k and tan - 1 x + cos - 1 y = π 2 possess solution and find all the solutions.

Equation-I : tan - 1 x + cos - 1 y = π 2 So, tan - 1 x = π 2 - cos - 1 y tan - 1 x = sin - 1 y . . . i Equation-II : ( tan - 1 x ) 2 + ( cos - 1 y ) 2 = π 2 k . . . ii Putting ( i ) in ( ii ) and using cos - 1 y = π 2 - sin - 1 y ( sin - 1 y ) 2 + π 2 - sin - 1 y 2 = π 2 k = ( sin - 1 y ) 2 + π 2 4 - 2 . π 2 . sin - 1 y + ( sin - 1 y ) 2 = π 2 k = 2 ( sin - 1 y ) - πsin - 1 y + π 2 1 4 - k = 0 sin - 1 y = π π 2 - 4 . 2 . π 2 1 4 - k 4 sin - 1 y = π ± π 8 k - 1 4 - π 2 ≤ sin - 1 y ≤ π 2 , so, permissible values of integral k are for k ≤ 0 , 8 k - 1 < 0 → not defined For k = 1 , sin - 1 y = π ± π 7 4 But π + π 7 4 > π 2 So, sin - 1 y = π + π 7 4 is not possible Hence sin - 1 y = π 4 ( 1 - 7 ) So, cos - 1 y = π 2 - π 4 ( 1 - 7 ) = cos - 1 y = π 4 ( 1 + 7 ) So, tan - 1 x = π 2 - cos - 1 y = π 4 ( 1 - 7 )

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-4

Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 4: Exercise-4

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 4: Exercise-4

Attempt the free practice questions on Chapter 14: Inverse Trigonometric Functions, Exercise 4: Exercise-4 with hints and solutions to strengthen your understanding. Alpha Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 4: Exercise-4 with Hints & Solutions

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-4>Q 3

Express $\cot ({cosec}^{- 1} x)$ as an algebraic function of $x .$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-4>Q 4

Express $\sin^{- 1} x$ in terms of
(i) $\cos^{- 1} \sqrt{1 - x^{2}}$
(ii) $\tan^{- 1} \frac{x}{\sqrt{1 - x^{2}}}$
(iii) $\cot^{- 1} \frac{\sqrt{1 - x^{2}}}{x}$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-4>Q 5

$\sin^{- 1} (\frac{x^{2}}{4} + \frac{y^{2}}{9}) + \cos^{- 1} (\frac{x}{2 \sqrt{2}} + \frac{y}{3 \sqrt{2}} - 2)$ equals to

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-4>Q 6

If $α = 2 \tan^{- 1} (\frac{1 + x}{1 - x}) & β = \sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})$ for $0 < x < 1,$ then prove that $α + β = π .$ What the value of $α + β$ will be if $x > 1 ?$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-4>Q 7

Solve $\{\cos^{- 1} x\} + [\tan^{- 1} x] = 0$ for real values of $x$ , where ${.}$ and $[.]$ are fractional part and greatest integer functions respectively.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-4>Q 8

Find the solution of $\sin^{- 1} \sqrt{\frac{x}{1 + x}} - \sin^{- 1} \frac{x - 1}{x + 1} = \sin^{- 1} \frac{1}{\sqrt{1 + x}}$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-4>Q 9

$(i)$ Find all positive integral solutions of the equation, $\tan^{- 1} x + \cot^{- 1} y = \tan^{- 1} 3$
$(ii)$ If $' k'$ be a positive integer, then show that the equation:
$\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} k$ has no non-zero integral solution.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-4>Q 10

Determine the integral values of $' k'$ for which the system ${(\tan^{- 1} x)}^{2} + {(\cos^{- 1} y)}^{2} = π^{2} k$ and
$\tan^{- 1} x + \cos^{- 1} y = \frac{π}{2}$ possess solution and find all the solutions.

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