If tan theta + cot theta = 5, then tan^2 thet | vedclass

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(D) Given that $	an 	heta + \cot 	heta = 5.$Squaring both sides of the equation,we get:$(	an 	heta + \cot 	heta)^2 = 5^2$Using the algebraic ide

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$23$

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(D) Given that $	an 	heta + \cot 	heta = 5.$Squaring both sides of the equation,we get:$(	an 	heta + \cot 	heta)^2 = 5^2$Using the algebraic identity $(a + b)^2 = a^2 + b^2 + 2ab$,we have:$	an^2 	heta + \cot^2 	heta + 2 	an 	heta \cot 	heta = 25$Since $	an 	heta \cot 	heta = 1$,the equation becomes:$	an^2 	heta + \cot^2 	heta + 2(1) = 25$$	an^2 	heta + \cot^2 	heta = 25 - 2$$	an^2 	heta + \cot^2 	heta = 23$

If $\tan \theta + \cot \theta = 5,$ then $\tan^2 \theta + \cot^2 \theta$ is

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