Find the derivative of tan^-1 x

Mathematics

1 answer:

k0ka [10]2 years ago

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Prove the following identity : Sin α . Cos α . Tan α = (1 – Cos α) (1 + Cos α)

Umnica [9.8K]

Let's work on the left side first. And remember that
the tangent is the same as sin/cos.

sin(a) cos(a) tan(a)

Substitute for the tangent:

[ sin(a) cos(a) ] [ sin(a)/cos(a) ]

Cancel the cos(a) from the top and bottom, and you're left with

[ sin(a) ] . . . . . [ sin(a) ] which is [ sin²(a) ] That's the left side.

Now, work on the right side:

[ 1 - cos(a) ] [ 1 + cos(a) ]

Multiply that all out, using FOIL:

[ 1 + cos(a) - cos(a) - cos²(a) ]

= [ 1 - cos²(a) ] That's the right side.

Do you remember that for any angle, sin²(b) + cos²(b) = 1 ?
Subtract cos²(b) from each side, and you have sin²(b) = 1 - cos²(b) for any angle.

So, on the right side, you could write [ sin²(a) ] .

Now look back about 9 lines, and compare that to the result we got for the left side .

They look quite similar. In fact, they're identical. And so the identity is proven.

Whew !

4 0

2 years ago

Read 2 more answers

Least common denominator?

Over [174]

Answer:

$\large\boxed{(s-2)(s+2)(s-3)=(s^2-4)(s-3)=s^3-3s^2-4s+12}$

Step-by-step explanation:

$s^2-4=s^2-2^2=(s-2)(s+2)\\\\\text{used}\ a^2-b^2=(a-b)(a+b)\\\\s^2-5s+6=s^2-2s-3s+6=s(s-2)-3(s-2)=(s-2)(s-3)\\\\\text{Therefore}\ LCD(s^2-4,\ s^2-5s+6)=(s-2)(s+2)(s-3)$