Trigonometric Ratios And Identities Question 1
Question 1 - 2024 (01 Feb Shift 1)
If $\tan \mathrm{A}=\frac{1}{\sqrt{x\left(x^{2}+x+1\right)}}, \tan B=\frac{\sqrt{x}}{\sqrt{x^{2}+x+1}}$ and $\tan C=\left(x^{-3}+x^{-2}+x^{-1}\right)^{\frac{1}{2}}, 0<A, B, C<\frac{\pi}{2}$, then $\mathrm{A}+\mathrm{B}$ is equal to :
(1) C
(2) $\pi-C$
(3) $2 \pi-C$
(4) $\frac{\pi}{2}-C$
Show Answer
Answer (1)
Solution
Finding $\tan (A+B)$ we get
$$ \begin{aligned} & \Rightarrow \tan (\mathrm{A}+\mathrm{B})= \ & \frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\frac{1}{\sqrt{x\left(x^{2}+x+1\right)}}+\frac{\sqrt{x}}{\sqrt{x^{2}+x+1}}}{1-\frac{1}{x^{2}+x+1}} \ & \Rightarrow \tan (\mathrm{A}+\mathrm{B})=\frac{(1+x)\left(\sqrt{x^{2}+x+1}\right)}{\left(x^{2}+x\right)(\sqrt{x})} \ & \frac{(1+x)\left(\sqrt{x^{2}+x+1}\right)}{\left(x^{2}+x\right)(\sqrt{x})} \ & \tan (A+B)=\frac{\sqrt{x^{2}+x+1}}{x \sqrt{x}}=\tan C \ & A+B=C \end{aligned} $$
