Trigonometric Equations
Solving Linear Trig Equation — Finding tanα
nta_pyq_2024_jan
Grade 11
Question:
If $\alpha$, $-\dfrac{\pi}{2}<\alpha<\dfrac{\pi}{2}$, is the solution of $4\cos\theta+5\sin\theta=1$, then the value of $\tan\alpha$ is
$\dfrac{10-\sqrt{10}}{6}$
$\dfrac{10-\sqrt{10}}{12}$
$\dfrac{\sqrt{10}-10}{12}$
$\dfrac{\sqrt{10}-10}{6}$
Step-by-Step Solution
Key Concept: Isolate $\cos\theta$: $4\cos\theta=1-5\sin\theta$. Square: $16(1-\sin^2\theta)=(1-5\sin\theta)^2\Rightarrow16-16\sin^2\theta=1-10\sin\theta+25\sin^2\theta\Rightarrow41\sin^2\theta-10\sin\theta-15=0$... Actually using $t=\tan\theta$: $4(1-t^2)/(1+t^2)+5\cdot2t/(1+t^2)=1\Rightarrow24t^2-40t-15=0$... solution gives $\tan\theta=(-10+\sqrt{10})/12$... wait: squaring gives $24\tan^2\theta+40\tan\theta+15=0\Rightarrow\tan\theta=(-10\pm\sqrt{10})/12$.
$\tan\alpha=\frac{\sqrt{10}-10}{12}$.
Correct Answer: 3