If sec theta + tan theta = m (m 1),then the | vedclass

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(B) Given: $\sec 	heta + 	an 	heta = m$ ....$(i)$We know the identity: $\sec^{2} 	heta - 	an^{2} 	heta = 1$Using $(a^{2} - b^{2}) = (a - b)(a +

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$\frac{m^{2}-1}{m^{2}+1}$

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(B) Given: $\sec 	heta + 	an 	heta = m$ ....$(i)$We know the identity: $\sec^{2} 	heta - 	an^{2} 	heta = 1$Using $(a^{2} - b^{2}) = (a - b)(a + b)$,we get:$(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1$Substituting $(i)$ into this,we get:$(\sec 	heta - 	an 	heta) \cdot m = 1 \Rightarrow \sec 	heta - 	an 	heta = \frac{1}{m}$ ....$(ii)$Adding $(i)$ and $(ii)$:$2 \sec 	heta = m + \frac{1}{m} = \frac{m^{2} + 1}{m} \Rightarrow \sec 	heta = \frac{m^{2} + 1}{2m}$Since $\cos 	heta = \frac{1}{\sec 	heta}$,we have $\cos 	heta = \frac{2m}{m^{2} + 1}$.Subtracting $(ii)$ from $(i)$:$2 	an 	heta = m - \frac{1}{m} = \frac{m^{2} - 1}{m} \Rightarrow 	an 	heta = \frac{m^{2} - 1}{2m}$Now,$\sin 	heta = 	an 	heta \cdot \cos 	heta = \left( \frac{m^{2} - 1}{2m} ight) \cdot \left( \frac{2m}{m^{2} + 1} ight) = \frac{m^{2} - 1}{m^{2} + 1}$.

If $\sec \theta + \tan \theta = m$ $(m > 1)$,then the value of $\sin \theta$ is $(0^{\circ} < \theta < 90^{\circ})$.

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