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

Sagot :

[tex]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}[/tex]
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