JEE Advanced 2011 Paper 1Question 1
Let $P=\{\theta:\sin\theta-\cos\theta=\sqrt2\cos\theta\}$ and $Q=\{\theta:\sin\theta+\cos\theta=\sqrt2\sin\theta\}$ be two sets. Then
Trigonometric Ratio & Identities
Solution
Concept. Reduce each condition to a value of $\tan\theta$. Formulas used. (1) $\dfrac{1}{\sqrt2-1}=\sqrt2+1$Step 1 - Set $P$. $\sin\theta=(\sqrt2+1)\cos\theta$. Here $\cos\theta\ne0$ (else $\sin\theta=0$ too), so $\tan\theta=\sqrt2+1$. Step 2 - Set $Q$. $\cos\theta=(\sqrt2-1)\sin\theta$. Here $\sin\theta\ne0$, so $\tan\theta=\frac{1}{\sqrt2-1}=\sqrt2+1$ by (1). Step 3 - Compare. Both sets are exactly $\{\theta:\tan\theta=\sqrt2+1\}$, so $P=Q$. Answer: (D)