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jasenka [17]
3 years ago
10

Tan^2x+sec^2x=1 for all values of x. true or false?

Mathematics
2 answers:
amid [387]3 years ago
5 0

This is not true.

\tan^2x+\sec^2x=\dfrac{\sin^2x}{\cos^2x}+\dfrac1{\cos^2x}=\dfrac{1-\cos^2x+1}{\cos^2x}=2\sec^2x-1

2\sec^2x-1=1\implies\sec^2x=1\implies\sec x=\pm1\implies x=n\pi

where is n is any integer. So suppose we pick some value of x other than these, say x=\dfrac\pi4. Then

\tan^2\dfrac\pi4+\sec^2\dfrac\pi4=1+2=3\neq1

padilas [110]3 years ago
3 0

The trigonometrical identity says : 1+tan^2x = sec^2 x that means

sec^2 x - tan^2 x = 1

However in our question, its tan^2 x + sec^2 x or say sec^2 x + tan^2 x = 1 which is completely against the identity.

Hense answer is False.

Additional information: There is, with a little difference, as equation which is true ie tan^2 x + sec x = 1 for some values of x.

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