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Question Number 185647 by Mingma last updated on 24/Jan/23

Answered by mahdipoor last updated on 24/Jan/23

$$secx+{tan}^{3}x.cscx=\frac{1}{cosx}+\frac{{sin}^{2}x}{{cos}^{3}x}=\frac{{cos}^{2}x+{sin}^{2}x}{{cos}^{3}x}=$$$$\frac{1}{{cos}^{3}x}={(1+{tan}^{2}x)}^{3/2}={(2-e^2)}^{3/2}$$$$$$$$$$

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