If tan theta + sec theta = l,then prove that | vedclass

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(A) Given: $	an 	heta + \sec 	heta = l$ .....$(i)$We know the identity: $\sec^{2} 	heta - 	an^{2} 	heta = 1$.This can be written as $(\sec 	het

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(A) Given: $	an 	heta + \sec 	heta = l$ .....$(i)$We know the identity: $\sec^{2} 	heta - 	an^{2} 	heta = 1$.This can be written as $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1$.Substituting the value of $(\sec 	heta + 	an 	heta)$ from equation $(i)$:$(\sec 	heta - 	an 	heta) \cdot l = 1$$\sec 	heta - 	an 	heta = \frac{1}{l}$ .....$(ii)$Now,adding equation $(i)$ and equation $(ii)$:$(	an 	heta + \sec 	heta) + (\sec 	heta - 	an 	heta) = l + \frac{1}{l}$$2 \sec 	heta = \frac{l^{2} + 1}{l}$Dividing both sides by $2$:$\sec 	heta = \frac{l^{2} + 1}{2l}$.Hence,it is proved.

If $\tan \theta + \sec \theta = l$,then prove that $\sec \theta = \frac{l^{2} + 1}{2l}$.

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