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 $^{\circ}$)

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