Prove that 1-tan32°tan13°-tan32°-tan13°=0

Mathematics

1 answer:

Step2247 [10]2 years ago

3 0

Answer:

Step-by-step explanation:

45=32+13

tan 45=tan(32+13)

$1=\frac{tan~ 32+tan~ 13}{1-tan ~32 ~tan~ 13} \\cross~multiply \\1-tan 32 ~tan~13=tan ~32+tan~13\\$

or 1-tan 32 tan 13-tan 32-tan 13=0

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How do you graph this f(x)= -3/2|x|+5 if x≤4

LenaWriter [7]

Answer:

see below

Step-by-step explanation:

I enter the equation into a graphing calculator and let it do the graphing.

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If you're graphing this by hand, you start by looking for the parent function. Here, it is |x|. That has a vertex of (0, 0) and a slope of +1 to the right of the vertex and a slope of -1 to the left of the vertex.

Here, the function is multiplied by -3/2, so will open downward and have slopes of magnitude 3/2 (not 1). The graph has been translated 5 units upward, so the vertex is (0, 5).

I'd start by plotting the vertex point at (0, 5), then identifying points with slope ±3/2 either side of it. To the left, it is left 2 and down 3 to (-2, 2). The points on the right of the vertex are symmetrically located about the y-axis, so one of them will be (2, 2).

Of course, you don't plot any function values for x > 4.