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Dovator [93]
3 years ago
5

If t is any real number, prove that 1+(tant)^2=(sect)^2

Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
6 0
1+\left(tant\right)^2=\left(sect\right)^2\\\\L=1+\left(\frac{sint}{cost}\right)^2=1+\frac{sin^2t}{cos^2t}=\frac{cos^2t}{cos^2t}+\frac{sin^2t}{cos^2t}=\frac{cos^2t+sin^2}{cos^2t}=\frac{1}{cos^2t}\\\\=\left(\frac{1}{cost}\right)^2=(sect)^2=R\\\\====================================\\\\\\tanx=\frac{sinx}{cosx}\\\\sin^2x+cos^2x=1\\\\secx=\frac{1}{cosx}
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Step-by-step explanation:

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