Question

Prove that $4tan^{-1}(\frac{1}{5}) + tan^{-1}(\frac{1}{99}) - tan^{-1}(\frac{1}{70}) = \frac{\pi}{4}$.

Answer 1

Solution :

i) Let tan^-1 ( 1/5 ) = A => tanA = 1/5

tan^-1 ( 1/99 ) = B => tanB = 1/99

tan^-1 ( 1/70 ) = C => tanC = 1/70

ii ) Tan 2A = ( 2tanA )/( 1 - tan²A )

= ( 2/5 )/[ 1 - ( 1/5 )² ]

= (2/5 )/[ 1 - 1/25 ]

= (2/5)/[ ( 25 - 1 )/25 ]

= (2/5)/(24/25)

= ( 2 × 25 )/( 5 × 24 )

= 5/12

iii ) tan 4A = ( 2tan2A )/( 1 - tan²2A )

= ( 2 × 5/12 )/[ 1 - ( 5/12 )² ]

= ( 5/6 )/( 1 - 25/144 )

= ( 5/6 )/[ (144 - 25 )/144 ]

= (5/6)/( 119/144)

= ( 5 × 144 )/( 6 × 119 )

= 120/119

iv ) tan( 4A + B )

= ( tan4A + tanB )/( 1 - tan4AtanB )

= [(120/119)+1/99]/[1-(120/119)(1/99)]

= [ 11880+119 ]/( 11781 - 120 )

= 11999/11661

v ) tan( 4A+B-C)

= [tan(4A+B)-tanC]/[1+tan(4A+B)tanC]

= [(11999/1166)-1/70]/[1+(11999/1166)(1/70)]

= 828269/828269

= 1

= tan ( π/4 )

Therefore ,

4A+B - C = π/4

4tan^-1(1/5)+tan^-1(1/99)-tan^-1(1/70)=π/4

••••

Answer 2

HELLO DEAR,

Answer:

Step-by-step explanation:

GIVEN:- 4tan-¹(1/5) + tan-¹(1/99) - tan-¹(1/70)

we know, [tan-¹x - tan-¹y = tan-¹ (x - y)/(1 + xy)]

=> 4tan-¹(1/5) + tan-¹ [1/99 - 1/70]/[1 + (1/99)*(1/70)]

=> 4tan-¹(1/5) + tan-¹ [(70 - 99)]/[6930 + 1]

=> 4tan-¹(1/5) + tan-¹[-29/6931]

=> 4tan-¹(1/5) - tan-¹[1/239]

=> 2[2tan-¹(1/5)] - tan-¹(1/239)

=> 2[tan-¹{(10/25)}/{(25 - 1)/25} - tan-¹(1/239)

=> 2tan-¹(5/12) - tan-¹(1/239)

=> tan-¹(120)/(144 - 25) - tan-¹(1/239)

=> tan-¹(120/119) - tan-¹(1/239)

=> tan-¹ {120/119 - 1/239}/{1 + (120/119)*(1/239)}

=> tan-¹ {(28680 - 119)/(28441 + 120)}

=> tan-¹ {28561/28561}

=> tan-¹(1) = π/4

I HOPE IT'S HELP YOU DEAR,

THANKS