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

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Question

(A) Given $\cot 	heta + 	an 	heta = m$ $\Rightarrow \frac{1}{	an 	heta} + 	an 	heta = m$ $\Rightarrow \frac{1 + 	an^2 	heta}{	an 	heta} = m

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $\cot 	heta + 	an 	heta = m$ $\Rightarrow \frac{1}{	an 	heta} + 	an 	heta = m$ $\Rightarrow \frac{1 + 	an^2 	heta}{	an 	heta} = m$ $\Rightarrow \sec^2 	heta = m 	an 	heta$ ... $(i)$ Given $\sec 	heta - \cos 	heta = n$ $\Rightarrow \frac{1}{\cos 	heta} - \cos 	heta = n$ $\Rightarrow \frac{1 - \cos^2 	heta}{\cos 	heta} = n$ $\Rightarrow \sin^2 	heta = n \cos 	heta$ $\Rightarrow 	an^2 	heta \cos^2 	heta = n \cos 	heta$ $\Rightarrow 	an^2 	heta \cos 	heta = n$ $\Rightarrow 	an^2 	heta = n \sec 	heta$ ... $(ii)$ From $(i)$,$	an 	heta = \frac{\sec^2 	heta}{m}$. Substituting in $(ii)$: $(\frac{\sec^2 	heta}{m})^2 = n \sec 	heta$ $\Rightarrow \frac{\sec^4 	heta}{m^2} = n \sec 	heta$ $\Rightarrow \sec^3 	heta = m^2 n$ $\Rightarrow \sec 	heta = (m^2 n)^{1/3}$ From $(i)$,$	an 	heta = \frac{\sec^2 	heta}{m} = \frac{(m^2 n)^{2/3}}{m} = (m n^2)^{1/3}$ Using $\sec^2 	heta - 	an^2 	heta = 1$: $(m^2 n)^{2/3} - (m n^2)^{2/3} = 1$ Multiplying by $m/m$ and $n/n$ respectively: $m \frac{(m^2 n)^{2/3}}{m} - n \frac{(m n^2)^{2/3}}{n} = 1$ $m (m n^2)^{1/3} - n (n m^2)^{1/3} = 1$.

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

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