Question

2tan⁻¹5 + tan⁻¹5/12=......,Select Proper option from the given options.
(a) π/4
(b) 2π/3
(c) π
(d) π/2

Answer 1

it is given that $2tan^{-1}5+tan^{-1}\frac{5}{12}$

Let $2tan^{-1}5=A$
tanA/2 = 5
tanA = 2tanA/2/(1 - tan²A/2) [ from formula]
= {2 × 5}/(1 - 5²)
= 10/-24
= -5/12
hence, tanA = -5/12 .......(1)

Let tan^-1(5/12) = B
then, tanB = 5/12 ........(2)

now, $2tan^{-1}5+tan^{-1}\frac{5}{12}$
= A + B

taking tangent both sides,
$tan(tan^{-1}5+tan^{-1}\frac{5}{12})$ = tan(A + B)
= (tanA + tanB)/(1 - tanA.tanB)

from equations (1) and (2),

= (5/12 - 5/12)/(1 - 5/12 × -5/12)
= 0

so, $tan(tan^{-1}5+tan^{-1}\frac{5}{12})$ = 0

$tan^{-1}5+tan^{-1}\frac{5}{12}=tan^{-1}0=\pi$

hence, option (c) is correct.

Answer 2

HELLO DEAR,



2tan-¹5 + tan-¹(5/12)

[as 2tan-¹x = tan-¹2*x/(1 - x²)]

tan-¹ 10/(1 - 25) + tan-¹ (5/12)

tan-¹ 10/(-24) + tan-¹ (5/12)

tan-¹ (5/-12) + tan-¹ (5/12)

[as tan-¹x + tan-¹y = tan-¹ (x + y)/(1 - xy)]

therefore,
tan-¹{ (5/12) + (-5/12) } / { 1 - (5/12)*(-5/12)}

tan-¹ {0/(144 + 5)/144}

tan-¹ 0 = 0 or π


HENCE, option (c) is correct,


I HOPE ITS HELP YOU DEAR,
THANKS