Question

Q1.if tan inverse 2 + tan inverse 3 + x = π then find the value of x.​

Answer 1

Step-by-step explanation:

Given,

$\displaystyle\small\longrightarrow \tan^{-1}(2)+\tan^{-1}(3)+x=\pi$

$Since \displaystyle\small\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left(\dfrac{a+b}{1-ab}\right),$

$\displaystyle\small\longrightarrow \tan^{-1}\left(\dfrac{2+3}{1-2\cdot3}\right)+x=\pi$

$\displaystyle\small\longrightarrow \tan^{-1}\left(-1\right)+x=\pi$

$We \: know \: \displaystyle\small\tan^{-1}(-1)=-\dfrac{\pi}{4}$

but here,

$\displaystyle\small\longrightarrow\tan^{-1}(2)\in\left(0,\ \dfrac{\pi}{2}\right)$

$\displaystyle\small\longrightarrow\tan^{-1}(3)\in\left(0,\ \dfrac{\pi}{2}\right)$

So,

$\displaystyle\small\longrightarrow \tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}(-1)\in(0,\ \pi)$

$\displaystyle\small\Longrightarrow\tan^{-1}(-1)\in\left(\dfrac{\pi}{2},\ \pi\right)$

$</p><p>\displaystyle\small\Longrightarrow\tan^{-1}(-1)=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}$

Therefore,

$\displaystyle\small\longrightarrow\dfrac{3\pi}{4}+x=\pi$

$\displaystyle\small\text{ \longrightarrow\underline{\underline{x=\dfrac{\pi}{4}}} }$

Answer 2

Step-by-step explanation:

Given :-

tan inverse 2 + tan inverse 3 + x = π

To find:-

Find the value of x ?

Solution:-

Given that :

Tan-¹ 2 + Tan-¹ 3 + x = π

Let a = 2

Let b = 3

ab = 2×3 = 6 > 1

ab > 1

We know that

Tan-¹ a+ Tan-¹ b = π + Tan-¹[(a+b)/(1-ab)]

On putting a = 2 and b = 3 in the above formula then

=>π + Tan-¹[ (2+3)/(1-(2)(3)] + X = π

=>π + Tan-¹[5/(1-6)] +X = π

=>π + Tan-¹(5/-5) +X = π

=> π + Tan-¹(-1) +X = π

=> π- 45° +X = π

=>-45° +X = π-π

=> -45° +X = 0

=> X = 45° or

=> X = 180°/4

=> X = π/4

Therefore, X = 45° or π/4

Answer:-

The value of x for the given problem is 45° or π/4

Used formulae:-

• If ab > 1 then,

Tan-¹ a+ Tan-¹ b = π + Tan-¹[(a+b)/(1-ab)]

• π = 180°
• π/4 = 45°
• Tan 45° = 1
• Tan (-A) = -Tan A