Question

The value of $\theta + \varphi = \frac{\pi}{4}$lying between 0 and $\tan(\pi\cos\theta) = \tan\left( \frac{\pi}{2} - \pi\sin\theta \right)$and satisfying the equation

$\therefore$.

• A. $\sin\theta + \cos\theta = \frac{1}{2}$ or $\Rightarrow \cos\left( \theta - \frac{\pi}{4} \right) = \frac{1}{2\sqrt{2}}$
• B. $\tan(\pi\cos\theta) = \tan\left( \frac{\pi}{2} - \pi\sin\theta \right)$
• C. $\therefore\sin\theta + \cos\theta = \frac{1}{2}$
• D. None of these

Answer 1

Answer: (A)

$\sin\theta + \cos\theta = \frac{1}{2}$ or $\Rightarrow \cos\left( \theta - \frac{\pi}{4} \right) = \frac{1}{2\sqrt{2}}$

Answer 2

The given determinant

(Applying $\left( \frac{a}{s - a} \right)\left( \frac{b(s - c) - c(s - b)}{(s - b)(s - c)} \right)$ and $\left( \frac{c}{s - c} \right)\left( \frac{a(s - b) - b(s - a)}{(s - a)(s - b)} \right)$) reduces to

$\frac{1}{c} + \frac{1}{a} = \frac{2}{b}$

$a,b,c(a^{2} + b^{2} - 2ab)\cos^{2}\frac{C}{2} + (a^{2} + b^{2} + 2ab)\sin^{2}\frac{C}{2}$

( By expanding along $= a^{2} + b^{2} + 2ab\left( \sin^{2}\frac{C}{2} - \cos^{2}\frac{C}{2} \right)$

⇒ $2s = a + b + c$ ⇒ $\cos\frac{B}{2} = \sqrt{\frac{30 \times 6}{320}} = \frac{3}{4}$

⇒ $\cos A + \cos C = 4\sin^{2}\frac{1}{2}B$ or $2\cos\frac{A + C}{2}\cos\frac{A - C}{2} = 4\sin^{2}\frac{B}{2}$, ($\cos\frac{A + C}{2}\cos\frac{A - C}{2} = 2\sin^{2}\frac{B}{2}$)

Since, $\cos\left( \frac{A - C}{2} \right) = 2\sin\frac{B}{2}$ ⇒ $\cos\frac{A}{2}\cos\frac{C}{2} + \sin\frac{A}{2}\sin\frac{C}{2} = 2\sin\frac{B}{2}$ ⇒ $\sqrt{\frac{s(s - a)}{bc}}\sqrt{\frac{s(s - c)}{ab}} + \sqrt{\frac{(s - b)(s - c)}{bc}}\sqrt{\frac{(s - a)(s - b)}{ab}}$.