Trigonometry & Inverse Trigonometry
Algebraic identities for sec²(tan⁻¹x) and cosec²(cot⁻¹x)
nta_pyq_2025_apr
Grade 12
Question:
If for some $\alpha,\beta$; $\alpha \leq \beta$, $\alpha+\beta = 8$ and $\sec^2(\tan^{-1}\alpha)+\operatorname{cosec}^2(\cot^{-1}\beta) = 36$, then $\alpha^2+\beta$ is ________.
Step-by-Step Solution
Key Concept: Use $\sec^2(\tan^{-1}x)=1+x^2$ and $\operatorname{cosec}^2(\cot^{-1}x)=1+x^2$ to convert the equation to $\alpha^2+\beta^2=34$, then combine with $\alpha+\beta=8$.
Let $\tan^{-1}\alpha=A$, $\cot^{-1}\beta=B$. Then $\sec^2 A+\operatorname{cosec}^2 B=(1+\alpha^2)+(1+\beta^2)=36 \Rightarrow \alpha^2+\beta^2=34$.
Given $\alpha+\beta=8$: $(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta=64 \Rightarrow 34+2\alpha\beta=64 \Rightarrow \alpha\beta=15$.
$\alpha,\beta$ are roots of $x^2-8x+15=0 \Rightarrow (x-3)(x-5)=0$, so $x=3$ or $5$.
Since $\alpha\leq\beta$: $\alpha=3,\,\beta=5$.
Therefore $\alpha^2+\beta=9+5=14$.
Correct Answer: 14