If theta is an acute angle and tan theta + co | vedclass

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

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$2$

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(B) Given the equation: $	an 	heta + \cot 	heta = 2$.Since $\cot 	heta = \frac{1}{	an 	heta}$,we can write the equation as $	an 	heta + \frac{1}{	an 	heta} = 2$.Let $x = 	an 	heta$. Then $x + \frac{1}{x} = 2$,which simplifies to $x^2 - 2x + 1 = 0$.This is $(x - 1)^2 = 0$,which gives $x = 1$.Therefore,$	an 	heta = 1$,which implies $	heta = 45^{\circ}$ (since $	heta$ is an acute angle).Now,we need to find the value of $	an^{5} 	heta + \cot^{5} 	heta$.Substituting $	an 	heta = 1$ and $\cot 	heta = 1$:$1^{5} + 1^{5} = 1 + 1 = 2$.

If $\theta$ is an acute angle and $\tan \theta + \cot \theta = 2$,then the value of $\tan^{5} \theta + \cot^{5} \theta$ is

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