If cot theta + tan theta = m and sec theta - | vedclass

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(A) Given: $\cot 	heta + 	an 	heta = m$ and $\sec 	heta - \cos 	heta = n$.$1$. From $\cot 	heta + 	an 	heta = m$:$\frac{1}{	an 	heta} + 	an

Vedclass · Accepted Answer

$m(mn^2)^{1/3} - n(nm^2)^{1/3} = 1$

Vedclass · Answer

(A) Given: $\cot 	heta + 	an 	heta = m$ and $\sec 	heta - \cos 	heta = n$.$1$. From $\cot 	heta + 	an 	heta = m$:$\frac{1}{	an 	heta} + 	an 	heta = m \implies \frac{1 + 	an^2 	heta}{	an 	heta} = m \implies \sec^2 	heta = m 	an 	heta$  --- $(i)$$2$. From $\sec 	heta - \cos 	heta = n$:$\sec 	heta - \frac{1}{\sec 	heta} = n \implies \frac{\sec^2 	heta - 1}{\sec 	heta} = n \implies 	an^2 	heta = n \sec 	heta$ --- $(ii)$$3$. Substitute $\sec^2 	heta = m 	an 	heta$ into $(ii)$:$(	an^2 	heta)^2 = n^2 \sec^2 	heta \implies 	an^4 	heta = n^2 (m 	an 	heta) \implies 	an^3 	heta = n^2 m \implies 	an 	heta = (n^2 m)^{1/3}$.$4$. Similarly,find $\sec 	heta$:Since $\sec^2 	heta = m 	an 	heta = m(n^2 m)^{1/3} = (m^4 n^2)^{1/3}$,then $\sec 	heta = (m^2 n)^{1/3}$.$5$. Using the identity $\sec^2 	heta - 	an^2 	heta = 1$:$(\sec 	heta)^2 - (	an 	heta)^2 = 1$$(m^2 n)^{2/3} - (n^2 m)^{2/3} = 1$ is not the form,let's use $\sec^2 	heta = m 	an 	heta$ and $	an^2 	heta = n \sec 	heta$:$m 	an 	heta - n \sec 	heta = 1$ is not correct,rather substitute $	an 	heta = (n^2 m)^{1/3}$ and $\sec 	heta = (m^2 n)^{1/3}$ into $\sec^2 	heta - 	an^2 	heta = 1$ is not the path. Actually,$m 	an 	heta - n \sec 	heta$ is not $1$. Using $\sec^2 	heta = m 	an 	heta$ and $	an^2 	heta = n \sec 	heta$,we have $\sec^2 	heta - 	an^2 	heta = 1 \implies (m 	an 	heta) - (n \sec 	heta)$ is not $1$. Wait,$\sec^2 	heta = m 	an 	heta$ and $	an^2 	heta = n \sec 	heta$. Then $\sec^2 	heta = m(n^2 m)^{1/3} = m^{4/3} n^{2/3}$ and $	an^2 	heta = n(m^2 n)^{1/3} = n^{4/3} m^{2/3}$.Thus,$m(mn^2)^{1/3} - n(nm^2)^{1/3} = 1$.

If $\cot \theta + \tan \theta = m$ and $\sec \theta - \cos \theta = n,$ then which of the following is correct?

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