If tanA=1√(xx^2+x+1),tanB=√x√(x^2+x+1) and tanC=x−3+x−2+x−1^12,0<A,B,C<(π)/2, then A+B is equal to:

Correct answer: (A) C. Given, tanA=(1√(xx^2+x+1),tanB=√x√(x^2+x+1) and tanC=x−3+x−2+x−1^12=√(1+x+x^2)x√xWe know that,tanA+B=tanA+tanB1−tanAtanB⇒tanA+B=1√(xx^2+x+1)+√x√(x^2+x+1)1−1)/(x^2+x+1) ⇒tanA+B=(1+x√(x^2+x+1))/(x^2+x√x) ⇒tanA+B=(1+x√(x^2+x+1))/(x^2+x√x) ⇒tanA+B=√(x^2+x+1)x√x ⇒tanA+B=tanC ⇒A+B=C

Q38 / 90 · JEE Main 2024 (01 Feb Shift 1) · Mathematics

If $\tan A = \frac{1}{\sqrt{x (x^{2} + x + 1)}}, \tan B = \frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}$ and $\tan C = {(x^{- 3} + x^{- 2} + x^{- 1})}^{\frac{1}{2}}, 0 < A, B, C < \frac{π}{2}$ , then $A + B$ is equal to:

Correct answer
A
$C$
B
$π - C$
C
$2 π - C$
D
$\frac{π}{2} - C$

Solution

Given,

\tan A = \frac{1}{\sqrt{x (x^{2} + x + 1)}}, \tan B = \frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}

and

\tan C = {(x^{- 3} + x^{- 2} + x^{- 1})}^{\frac{1}{2}} = \frac{\sqrt{1 + x + x^{2}}}{x \sqrt{x}}

We know that,

$\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

$\Rightarrow \tan (A + B) = \frac{\frac{1}{\sqrt{x (x^{2} + x + 1)}} + \frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}}{1 - \frac{1}{x^{2} + x + 1}}$

$\Rightarrow \tan (A + B) = \frac{(1 + x) (\sqrt{x^{2} + x + 1})}{(x^{2} + x) (\sqrt{x})}$

$\Rightarrow \tan (A + B) = \frac{\sqrt{x^{2} + x + 1}}{x \sqrt{x}}$

$\Rightarrow \tan (A + B) = \tan C$

$\Rightarrow A + B = C$

Hence, the correct answer is option: 'A'.

If tanA=1xx2+x+1,tanB=xx2+x+1 and tanC=x−3+x−2+x−112,0<A,B,C<π2, then A+B is equal to:

Solution

Relevant Questions

If $\tan A = \frac{1}{\sqrt{x (x^{2} + x + 1)}}, \tan B = \frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}$ and $\tan C = {(x^{- 3} + x^{- 2} + x^{- 1})}^{\frac{1}{2}}, 0 < A, B, C < \frac{π}{2}$ , then $A + B$ is equal to: