If tan theta + cot theta = 2 and 0^{circ}

Question

(D) Given the equation: $	an 	heta + \cot 	heta = 2$.Since $\cot 	heta = \frac{1}{	an 	heta}$,we can write: $	an 	heta + \frac{1}{	an 	heta}

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(D) Given the equation: $	an 	heta + \cot 	heta = 2$.Since $\cot 	heta = \frac{1}{	an 	heta}$,we can write: $	an 	heta + \frac{1}{	an 	heta} = 2$.Let $x = 	an 	heta$. Then $x + \frac{1}{x} = 2$.Multiplying by $x$,we get $x^2 + 1 = 2x$,which simplifies to $x^2 - 2x + 1 = 0$.This is a perfect square: $(x - 1)^2 = 0$.Therefore,$x = 1$,which means $	an 	heta = 1$.Since $0^{\circ} < 	heta < 90^{\circ}$,$	an 	heta = 1$ implies $	heta = 45^{\circ}$.

If $\tan \theta + \cot \theta = 2$ and $0^{\circ} < \theta < 90^{\circ}$,then the value of $\theta$ is (in $^{\circ}$)

Similar Questions

For any $\theta \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right)$,the expression $3(\sin \theta - \cos \theta)^4 + 6(\sin \theta + \cos \theta)^2 + 4\sin^6 \theta$ equals

$\tan 75^\circ - \cot 75^\circ = $

If $\sin \theta = -\frac{4}{5}$ and $\theta$ lies in the third quadrant,then $\cos \frac{\theta}{2} = $

The maximum value of $3\cos \theta - 4\sin \theta $ is

The exact value of $\cos \frac{2\pi}{28} \csc \frac{3\pi}{28} + \cos \frac{6\pi}{28} \csc \frac{9\pi}{28} + \cos \frac{18\pi}{28} \csc \frac{27\pi}{28}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module