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Dovator [93]
4 years ago
5

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

Mathematics
1 answer:
BlackZzzverrR [31]4 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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Question 7 of 10
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The correct answer is option D which is 0.798

The complete question is given below:-

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d.) 0.798

<h3>What is an equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

Since we know that an exponential function is in the form:y= ab.^x where,

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Upon looking at our given choices we can see that options a, b and c are in form 1+r, while the choice provided in option d is less than 1 and in form 1-r, therefore, option d is the correct choice.

Therefore the correct answer is option D which is 0.798

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