Question on: WAEC Mathematics - 2010
Given that tan x = 1, where 0o \(\geq\) x 90o, evaluate \(\frac{1 - \sin^2 x}{\cos x}\)
A
2\(\sqrt{2}\)
B
\(\sqrt{2}\)
C
\(\frac{\sqrt{2}}{2}\)
D
\(\frac{1}{2}\)
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Correct Option: C
Given tan x = 1
x = tan-1(1)
x = 45o
Now, \(\frac{1 - ( \frac{1}{\sqrt{2} )^2}}{\frac{1}{\sqrt{2}}}\)
= \(\frac{1 - \frac{1}{2}}{\frac{1}{2}}\)
= \(\frac{1}{2} + \frac{1}{\sqrt{2}}\)
= \(\frac{1}{2} \times \frac{1}{\sqrt{2}}\)
= \(\frac{\sqrt{2}}{2}\)
x = tan-1(1)
x = 45o
Now, \(\frac{1 - ( \frac{1}{\sqrt{2} )^2}}{\frac{1}{\sqrt{2}}}\)
= \(\frac{1 - \frac{1}{2}}{\frac{1}{2}}\)
= \(\frac{1}{2} + \frac{1}{\sqrt{2}}\)
= \(\frac{1}{2} \times \frac{1}{\sqrt{2}}\)
= \(\frac{\sqrt{2}}{2}\)
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