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2 years ago

if the terminal side of angle theta passes through point (6,-8) what is the value of six trigonometric functions

Mathematics

1 answer:

qaws [65]2 years ago

7 0

Asquare+bsquare=csquare
sintheta=-4/5
csctheta=-5/4
costheta=3/5
sectheta=5/4
tantheta=-4/3
cottheta=-3/4

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Matrices A and B are square matrices of the same size. Prove Tr(c(A + B)) = C (Tr(A) + Tr(B)).

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Answer with Step-by-step explanation:

We are given that two matrices A and B are square matrices of the same size.

We have to prove that

Tr(C(A+B)=C(Tr(A)+Tr(B))

Where C is constant

We know that tr A=Sum of diagonal elements of A

Therefore,

Tr(A)=Sum of diagonal elements of A

Tr(B)=Sum of diagonal elements of B

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C(Tr(B))= $C\cdot$ Sum of diagonal elements of B

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Tr(C(A+B)=Sum of diagonal elements of (C(A+B))

Suppose ,A= $\left[\begin{array}{ccc}1&0\\1&1\end{array}\right]$

B= $\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]$

Tr(A)=1+1=2

Tr(B)=1+1=2

C(Tr(A)+Tr(B))=C(2+2)=4C

A+B= $\left[\begin{array}{ccc}1&0\\1&1\end{array}\right]+\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]$

A+B= $\left[\begin{array}{ccc}2&1\\2&2\end{array}\right]$

C(A+B)= $\left[\begin{array}{ccc}2C&C\\2C&2C\end{array}\right]$

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Hence, Tr(C(A+B)=C(Tr(A)+Tr(B))

Hence, proved.

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