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koban [17]
2 years ago
6

Sec (90 -A) sinA = cot (90-A). tan( 90- A)​

Mathematics
2 answers:
Lapatulllka [165]2 years ago
8 0

L.H.S=sec(90-A)·sinA

        =cosecA·sinA ;[sec(90-A)= cosecA]

        =1/sinA·sinA ;[cosecA=1/sinA]

        =1

R.H.S=cot(90-A)·tan(90-A)

        =tanA·cotA ;[cot(90-A)=tanA, tan(90-A)=cotA]

        =tanA·1/tanA ;[cotA=1/tanA]

        =1

thus, L.H.S=R.H.S

[Proved]

djyliett [7]2 years ago
3 0

Step-by-step explanation:

sec(90-A) . Sin A = cot (90-A) . tan(90-A)

cosec X sinA = tanA X cotA

1/sinA X sinA = tanA X 1/tanA

1=1

Hence proved

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s344n2d4d5 [400]

Answer:

A. H(x) is an inverse of F(x)

Step-by-step explanation:

The given functions are:

F(x)=\sqrt{x-2}

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We compose F(x) and G(x) to get:

(F\circ G)(x)=F(G(x))

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(F\circ G)(x)=\sqrt{(x-2)^2-2}

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(F\circ G)(x)=\sqrt{x^2-4x+2}

(F\circ G)(x)\ne x

Hence G(x) is not an inverse of F(x).

We now compose H(x) and G(x).

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3 years ago
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Leno4ka [110]
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So I would say A. Thanks for asking your question today and have a good day on Brainly.com! :)
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