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

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Trigonometric Ratios and Identities

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

Answer: (A)

tan A = \frac{1}{x x ^{2} + x + 1}, tan B = \frac{x}{x ^{2} + x + 1}, tan C = \frac{x ^{2} + x + 1}{x x}

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

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

tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B} = \frac{( 1 + x )}{x x ^{2} + x + 1} \cdot \frac{x ^{2} + x + 1}{x ( x + 1 )} = \frac{x ^{2} + x + 1}{x x} = tan C

Since

0 < A, B, C < π /2

and

A + B

stays within the principal range for these expressions,

A + B = C

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option