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salantis [7]
2 years ago
5

What basic trigonometric identity would you use to verify that sinx cosx tanx=1-cos^2x

Mathematics
2 answers:
professor190 [17]2 years ago
7 0
You could use the following identities in verifying the trigonometric expression <span>sinx cosx tanx=1-cos^2x:

</span><span>tanx = sinx/cosx
sin^2x+cos^2x = 1
</span>
When those identities are used, you could now verify the equation.

Hope that helps you.
andre [41]2 years ago
5 0

Answer:

sin^2x+cos^2x=1 is used to prove the given equation.

Step-by-step explanation:

We have given an equation:

sinx\cdot cosx\cdot tanx=1-cos^2x

We will consider right hand side first

tan x can be written as:

tanx=\frac{sinx}{cosx}

So, on substituting tan x in the left hand side of the given equation we get:

sinx\cdot cosx\cdot \frac{sinx}{cosx}

Cancel common term from denominator and numerator which is cosx we get

sin^2x

And we have an identity:

sin^2x+cos^2x=1

\Rightarrow sin^2x=1-cos^2x

Hence, right hand side is equal to sin^2x.


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Answer:

<u>It</u><u> </u><u>is</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>4</u><u>x</u><u> </u><u>+</u><u> </u><u>1</u><u>0</u>

Step-by-step explanation:

Let's first get the slope of the curve.

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introduce dy/dx :

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make dy/dx the subject:

2x - y -  \frac{dy}{dx}  + 2y \frac{dy}{dx}  = 0 \\  \\ 2y \frac{dy}{dx}  -  \frac{dy}{dx}  = y - 2x \\  \\  \frac{dy}{dx} (2y - 1) = y - 2x \\  \\  \frac{dy}{dx}  =  \frac{y - 2x}{2y - 1}

At point (-1, 2):

\frac{dy}{dx}  =  \frac{2 - 2( - 1)}{2(2) - 1}  \\  \\ slope =  \frac{4}{3}

but a tangent has the same slope as the curve:

y = mx + c

m is the slope

c is the y-intercept

At (-1, 2):

2 = ( - 1 \times  \frac{4}{3} ) + c \\  \\ c = 2 +  \frac{4}{3}  \\  \\ c =  \frac{10}{3}

equation:

y =  \frac{4}{3} x +  \frac{10}{3}  \\  \\ { \boxed{3y = 4x + 10}}

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