Question

sec²x+tan²x= a then what is the value of a
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Answer 1

ANSWER

First, convert [math]sec^2 (x)-tan^2 (x)[/math]into [math]\frac{[/math][math]1}{ cos^2 (x)} - \frac{ sin^2 (x)}{ [/math][math] cos^2 (x)}[/math]

Then you have [math] \frac {1-sin^2 (x)}{cos^2 (x)}[/math]

Then you use the formula of [math]1-sin^2 (x)=cos^2 (x)[/math]

And you just simplify it to 1, where cos(x) does not equal 0.

EDIT: Oops. How do I fix this?

Answer 2

Answer :-

${sec}^{2}x \: + {tan}^{2}x \: = \: a \\ = > {tan}^{2}x \: = 1 - {sec}^{2}x \\ = > {sec}^{2} x \: + 1 - {sec}^{2} x \\ = 1 \\ \\ value \: of \: a \: = 1$