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Viktor [21]
3 years ago
7

Simplify 1+sinx/cosx+cosx/1+sinx

Mathematics
1 answer:
Dmitry [639]3 years ago
6 0
\frac{1 + sin(x)}{cos(x)} + \frac{cos(x)}{1 + sin(x)}
\frac{[1 + sin(x)][1 +sin(x)]}{cos(x)[1 + sin(x)]} + \frac{cos(x)[cos(x)]}{cos(x)[1 + sin(x)]}
\frac{1 + 2sin(x) + sin^{2}(x)}{cos(x)[1 + sin(x)]} + \frac{cos^{2}(x)}{cos(x)[1 + sin(x)]}
\frac{1 + 2sin(x) + sin^{2}(x) + cos^{2}(x)}{cos(x)[1 + sin(x)]}
\frac{1 + 2sin(x) + 1}{cos(x)[1 + sin(x)]}
\frac{1 + 1 + 2sin(x)}{cos(x)[1 + sin(x)]}
\frac{2 + 2sin(x)}{cos(x)[1 + sin(x)]}
\frac{2[1 + sin(x)]}{cos(x)[1 + sin(x)]}
\frac{2}{cos(x)}
2sec(x)
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Answer:

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Step-by-step explanation:

• Let's first find Angle ACB

{ \rm{ \angle ACB + 23 \degree + 90 \degree = 180 \degree}} \\  \\ { \rm{ \angle ACB = (180 - 90 - 23) \degree}} \\  \\ { \underline{ \rm{ \:  \angle ACB = 67 \degree}}}

• From alternative angles, x = Angle BAC.

• Since AB = AC, then Angle ABC = Angle ACB

{ \rm{x +  \angle ABC +  \angle ACB = 180 \degree}} \\  \\ { \rm{x + 67 \degree + 67 \degree = 180 \degree}} \\  \\ { \rm{x + 134 \degree = 180 \degree}} \\  \\ { \boxed{ \rm{x = 46 \degree}}}

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Step-by-step explanation:

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Step-by-step explanation:

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-4

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Find the gradient of the line segment between the points (0,2) and (-2,10).​

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