Question

Which of the following statements is/are true?
Statement-I : If $cot\theta+tan\theta=2cosec\theta$, then the general value of $\theta$ is $n\pi \pm\frac{\pi}{3}$, $n \in Z$.
Statement-II : If $secx\,cos5x+1=0$, whrere $0 < x \le\frac{\pi}{2}$, then the value of $x$ is $\frac{\pi}{3}$.

• A. Only Statement-I
• B. Only Statement-II
• C. Both Statement-I and Statement-II
• D. Neither Statement-I nor Statement-II

Answer 1

Answer: (D)

Statement-I : Given, $cot\theta+tan\theta=2cosec\theta$
$\Rightarrow \frac{cos\,\theta}{sin\,\theta}+\frac{sin\,\theta}{cos\,\theta}=\frac{2}{sin\,\theta}$
$\Rightarrow \frac{cos^{2}\,\theta+sin^{2}\,\theta}{sin\,\theta\cdot cos\,\theta}=\frac{2}{sin\,\theta}$
$\Rightarrow \frac{1}{cos\,\theta}=2$
$\Rightarrow cos\,\theta=\frac{1}{2}$
$\Rightarrow cos\,\theta=cos \frac{\pi}{3}$
$\Rightarrow \theta=2n\pi \pm \frac{\pi}{3}$, where $n \in Z$

Statement-II : Given, $secx\, cos5x +1 = 0$
$\Rightarrow \frac{cos\,5x}{cos\,x}+1=0$
$\Rightarrow cos\,5x + cos\,x = 0$
$\Rightarrow 2\,cos\left(\frac{5x+x}{2}\right)\cdot cos\left(\frac{5x-x}{2}\right)=0$
$\Rightarrow 2cosx\cdot cos2x=0$
$\Rightarrow cos3x=0$ or $cos2x=0$
$\Rightarrow cos\,3x=cos \frac{\pi}{2}$ or $cos\,2x=cos \frac{\pi}{2}$
$\Rightarrow 3x=\frac{\pi}{2}$ or $2x=\frac{\pi}{2}$
$\Rightarrow x=\frac{\pi}{6} $or $x=\frac{\pi}{4}$
Hence, the solutions are $\frac{\pi}{4}$ and $\frac{\pi}{6}$