Prove that tan - 1 a ( a + b + c ) b c + tan - 1 b ( a + b + c ) a c + tan - 1 c ( a + b + c ) a b = π , where a , b , c are positive.

Let tan - 1 a ( a + b + c ) b c = A , tan - 1 b ( a + b + c ) a c = B tan - 1 c ( a + b + c ) a b = C where A , B , C ∈ 0 , π 2 . tan A = a ( a + b + c ) b c , tan B = b ( a + b + c ) a c , tan C = c ( a + b + c ) a b As we know that tan A + B + C = tan A + tan B + tan C - tan A tan B tan C 1 - tan A tan B - tan B tan C - tan A tan C From the above values tan A + tan B + tan C = a a + b + c + b a + b + c + c a + b + c a b c = a + b + c 3 2 a b c = tan A tan B tan C So tan ( A + B + C ) = 0 A + B + C = n π , n ∈ Z Since A + B + C ∈ 0 , 3 π 2 . A + B + C = π

Evaluate: cos - 1 x + cos - 1 x 2 + 1 2 3 - 3 x 2 , x ∈ 1 2 , 1

cos - 1 x + cos - 1 x 2 + 1 2 3 - 3 x 2 , x ∈ 1 2 , 1 = cos - 1 x + cos - 1 x · 1 2 + 3 2 1 - x 2 = cos - 1 x + cos - 1 x · 1 2 + 1 - 1 2 2 1 - x 2 As we know that cos - 1 x - cos - 1 y = cos - 1 x y + 1 - x 2 1 - y 2 , 0 ≤ x ≤ y ≤ 1 - cos - 1 x y + 1 - x 2 1 - y 2 , 0 ≤ y ≤ x ≤ 1 As it is given that x ≥ 1 2 . So the given expression becomes cos - 1 x + cos - 1 1 2 - cos - 1 x = π 3

If sin - 1 x a + sin - 1 y b = sin - 1 c 2 a b , show that: b 2 x 2 + 2 x y a 2 b 2 - c 4 + a 2 y 2 = c 4 .

We have sin - 1 x a + sin - 1 y b = sin - 1 c 2 a b Using interconversion cos - 1 a 2 - x 2 a + cos - 1 b 2 - y 2 b = cos - 1 a 2 b 2 - c 4 a b cos - 1 a 2 - x 2 a · b 2 - y 2 b - 1 - a 2 - x 2 a 2 1 - b 2 - y 2 b 2 = cos - 1 a 2 b 2 - c 4 a b (using cos - 1 x + cos - 1 y = cos - 1 x y - 1 - x 2 1 - y 2 , - 1 ≤ x , y ≤ 1 & x + y ≥ 0 ) a 2 - x 2 · b 2 - y 2 a b - x a · y b = a 2 b 2 - c 4 a b a 2 - x 2 · b 2 - y 2 - x y = a 2 b 2 - c 4 a 2 - x 2 b 2 - y 2 = ( x y + a 2 b 2 - c 4 ) 2 a 2 b 2 - a 2 y 2 - b 2 x 2 + x 2 y 2 = x 2 y 2 + a 2 b 2 - c 4 + 2 x y a 2 b 2 - c 4 b 2 x 2 + 2 x y a 2 b 2 - c 4 + a 2 y 2 = c 4

Solve: θ = tan - 1 2 tan 2 θ - 1 2 sin - 1 3 sin 2 θ 5 + 4 cos 2 θ .

Let α = 1 2 sin - 1 3 sin 2 θ 5 + 4 cos 2 θ sin 2 α = 3 sin 2 θ 5 + 4 cos 2 θ sin 2 α = 3 2 tan θ 1 + tan 2 θ 5 + 4 1 - tan 2 θ 1 + tan 2 θ 2 tan α 1 + tan 2 α = 6 tan θ 9 + tan 2 θ tan θ = 3 tan α Now θ = tan - 1 2 tan 2 θ - α tan ( θ + α ) = 2 tan 2 θ tan θ + tan α 1 - tan θ · tan α = 2 tan 2 θ ∴ tan θ + tan θ 3 1 - tan θ · tan θ 3 = 2 tan 2 θ 4 tan θ 3 - tan 2 θ = 2 tan 2 θ 2 tan θ 2 - tan θ 3 - tan 2 θ = 0 ⇒ tan θ · ( tan θ - 1 ) tan 2 θ + tan θ - 2 = 0 ⇒ tan θ · ( tan θ - 1 ) · ( tan θ - 1 ) · ( tan θ + 2 ) = 0 tan θ = 0 ⇒ θ = n π , n ∈ Z tan θ = 1 ⇒ θ = n π + π 4 , n ∈ Z tan θ = - 2 ⇒ θ = n π + tan - 1 ( - 2 ) , n ∈ Z

Solve the equation: cos - 1 x 2 - 1 x 2 + 1 + sin - 1 2 x x 2 + 1 + tan - 1 2 x x 2 - 1 = 2 π 3

Case I : | x | ≤ 1 or x 2 < 1 We have; cos - 1 - 1 - x 2 1 + x 2 + sin - 1 2 x 1 + x 2 + tan - 1 - 2 x 1 - x 2 = 2 π 3 π - cos - 1 1 - x 2 1 + x 2 + sin - 1 2 x 1 + x 2 - tan - 1 2 x 1 - x 2 = 2 π 3 ⇒ π 3 = 2 tan - 1 x + 2 tan - 1 ( x ) - 2 tan - 1 ( x ) ⇒ 2 tan - 1 ( x ) = π 3 ⇒ tan - 1 ( x ) = π 6 ⇒ x = tan ( π / 6 ) = 1 3 Case II : When | x | > 1 Then cos - 1 1 - 1 / x 2 1 + 1 / x 2 + sin - 1 2 · 1 / x 1 + ( 1 / x ) 2 + tan - 1 2 · 1 / x 1 - ( 1 / x ) 2 = 2 π 3 ⇒ 2 tan - 1 1 x + 2 tan - 1 1 x + 2 tan - 1 1 x = 2 π 3 ⇒ tan - 1 ( 1 / x ) = π 9 ⇒ 1 x = tan π 9 ⇒ x = 1 tan π / 9 = cot 20 ° > 1 ∴ x = 1 3 and c o t 20 ° .

Reduce to the simplest form: tan - 1 x cos θ 1 - x sin θ - cot - 1 cos θ x - sin θ

Given expression is tan - 1 x cos θ 1 - x sin θ - cot - 1 cos θ x - sin θ . = tan - 1 x cos θ 1 - x sin θ - tan - 1 x - sin θ cos θ = tan - 1 x cos θ ( 1 - x sin θ ) - ( x - sin θ ) cos θ 1 + x · cos θ ( 1 - x sin θ ) · ( x - sin θ ) cos θ = tan - 1 x cos 2 θ - ( x - sin θ ) ( 1 - x sin θ ) ( 1 - x sin θ ) · cos θ + x cos θ ( x - sin θ ) = tan - 1 x 2 sin θ + sin θ - 2 x sin 2 θ - x cos 2 θ cos θ 1 - 2 x sin θ + x 2 = tan - 1 sin θ x 2 + 1 - 2 x sin θ cos θ 1 - 2 x sin θ + x 2 = tan - 1 ( tan θ ) = θ . Thus, simplest form of given expression is θ .

If ϕ = tan - 1 x 3 2 k - x and θ = tan - 1 2 x - k k 3 , prove that one value of ( ϕ - θ ) is 30 °

Given, ϕ = tan - 1 x 3 2 k - x ⇒ tan ϕ = x 3 2 k - x Also, θ = tan - 1 2 x - k k 3 ⇒ tan θ = 2 x - k k 3 Now we know that tan ( ϕ - θ ) = tan ϕ - tan θ 1 + tan ϕ tan θ Putting the values tan ϕ - θ = x 3 2 k - x - 2 x - k k 3 1 + x 3 2 k - x 2 x - k k 3 = 3 k x - ( 2 x - k ) ( 2 k - x ) 3 k ( 2 k - x ) + x 3 ( 2 x - k ) = 3 k x - 4 k x + 2 x 2 + 2 k 2 - k x 3 2 k 2 - k x + 2 x 2 - k x tan ϕ - θ = 2 k 2 - 2 k x + 2 x 2 3 2 k 2 - 2 k x + 2 x 2 = 1 3 ⇒ ϕ - θ = n π + π 6 , n ∈ Z Therefore, one of the value of ϕ - θ is 30 ° .

E M B I B E

Skills in Mathematics Trigonometry for JEE Main & Advanced>Chapter 4 - Inverse Trigonometric Functions>Exercise on Level

Amit M Agarwal Solutions for Chapter: Inverse Trigonometric Functions, Exercise 2: Exercise on Level

Author:Amit M Agarwal

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Inverse Trigonometric Functions, Exercise 2: Exercise on Level

Attempt the practice questions on Chapter 4: Inverse Trigonometric Functions, Exercise 2: Exercise on Level with hints and solutions to strengthen your understanding. Skills in Mathematics Trigonometry for JEE Main & Advanced solutions are prepared by Experienced Embibe Experts.

Questions from Amit M Agarwal Solutions for Chapter: Inverse Trigonometric Functions, Exercise 2: Exercise on Level with Hints & Solutions

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Trigonometry for JEE Main & Advanced>Chapter 4 - Inverse Trigonometric Functions>Exercise on Level>Q 5

Prove that $\tan^{- 1} \sqrt{\frac{a (a + b + c)}{b c}} + \tan^{- 1} \sqrt{\frac{b (a + b + c)}{a c}} + \tan^{- 1} \sqrt{\frac{c (a + b + c)}{a b}} = π$ , where $a, b, c$ are positive.

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Trigonometry for JEE Main & Advanced>Chapter 4 - Inverse Trigonometric Functions>Exercise on Level>Q 6

Evaluate:

$\cos^{- 1} x + \cos^{- 1} \{\frac{x}{2} + \frac{1}{2} \sqrt{3 - 3 x^{2}}\}, x \in [\frac{1}{2}, 1]$

HARD

JEE Main

IMPORTANT

Skills in Mathematics Trigonometry for JEE Main & Advanced>Chapter 4 - Inverse Trigonometric Functions>Exercise on Level>Q 7

If $\sin^{- 1} \frac{x}{a} + \sin^{- 1} \frac{y}{b} = \sin^{- 1} \frac{c^{2}}{a b}$ , show that: $b^{2} x^{2} + 2 x y \sqrt{a^{2} b^{2} - c^{4}} + a^{2} y^{2} = c^{4}$ .