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Sloan [31]
4 years ago
11

Simplify for a lot of points (1-sin^2x)(1-tan^2x)

Mathematics
1 answer:
notka56 [123]4 years ago
6 0
1/sin^2x-1/tan^2x=
1/sin^2x-1/ (sin^2x/cos^2x)<<sin tan= sin/cos>>
= 1/sin^2x- cos^2x / sin^2x
= (1- cos^2x) / sin^2x <<combining into a single fraction>>
sin^2 x / sin^2x <<since 1- cos^2 x sin^2 x

=1
this simplifies to 1.
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3 years ago
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saul85 [17]

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3 years ago
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tankabanditka [31]
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another quadratic equation

if you hahve
ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}

1x^2-6x-2=0
a=1
b=-6
c=-2

x=\frac{-(-6)+/- \sqrt{(-6)^{2}-4(1)(-2)} }{2(1)}
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aprox
x=6.31662 or -0.316625
4 0
3 years ago
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