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if tan^2θ=8 find the exact value of sec^2θ
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45
but how
\(\sec^2x = 1+\tan^2 x\)
Recall the Pythagorean identity:
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csc^2x = 1 + cot^2x too @hartnn ?
yes.
thanks!!!
|dw:1362979754948:dw| By the Pythagorean theorem, we know \(a^2+b^2=c^2\), or, in this case, \(\cos^2x+\sin^2x=1\). Divide through by \(\cos^2x\) to yield \(1+\frac{\sin^2x}{\cos^2x}=\frac1{\cos^2x}\) -- or, equivalently, \(1+\tan^2x=\sec^2x\)
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