Question

If cot $\sin\left( \theta + \frac{\pi}{4} \right) = \frac{1}{2\sqrt{2}}$then $R_{1} \rightarrow R_{1} - R_{3}$

• A. $\sin\theta = \frac{1}{2}$
• B. $\left| \begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & - 1 \\ \sin^{2}\theta & \cos^{2}\theta & 1 + 4\sin 4\theta \end{matrix} \right| = 0$
• C. $\Rightarrow$
• D. $1 + 4\sin 4\theta + \cos^{2}\theta + \sin^{2}\theta = 0$

Answer 1

Answer: (A)

$\sin\theta = \frac{1}{2}$

Answer 2

Given, cot $= 2\sqrt{\frac{(s - a)(s - c)}{ac}}$

⇒ $\frac{\sqrt{(s - a)(s - c)}}{ac} + \frac{s - b}{b}\sqrt{\frac{(s - c)\ (s - a)}{ac}}$

∴$= 2\sqrt{\frac{(s - a)(s - c)}{ac}}$=$\frac{s}{b} + \frac{s - b}{b} = 2$

$a + c = 2b$ = $a,b,c$.