Question

tanA÷1+secA+1+secA÷tana=​

Answer 1

Answer:

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Answer 2

Formula Applied :

$\bullet\ \; \sf sec^2\theta-tan^2\theta=1\\\\\to \sf sec^2\theta=1+tan^2\theta\\\\\sf \bullet\ \; tan\theta=\dfrac{sin\theta}{cos\theta}\\\\\sf \bullet\ \; sec\theta=\dfrac{1}{cos\theta}$

Solution :

$\sf \dfrac{tan\theta}{1+sec\theta}+\dfrac{1+sec\theta}{tan\theta}\\\\\to \sf \dfrac{tan^2\theta+(1+sec\theta)^2}{tan\theta(1+sec\theta)}\\\\\to \sf \dfrac{(tan^2\theta+1)+sec^2\theta+2sec\theta}{tan\theta(1+sec\theta)}\\\\\to \sf \dfrac{sec^2\theta+sec^2\theta+2sec\theta}{tan\theta(1+sec\theta)}\\\\\to \sf \dfrac{2sec^2\theta+2sec\theta}{tan\theta(1+sec\theta)}\\\\\to \sf \dfrac{2sec\theta(sec\theta+1)}{tan\theta(sec\theta+1)}\\\\\to \sf \dfrac{2sec\theta}{tan\theta}$

$\to \sf \dfrac{2}{cos\theta} \times \dfrac{cos\theta}{sin\theta}\\\\\to \sf \dfrac{2}{sin\theta}\\\\\leadsto \sf 2csc\theta$