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vladimir2022 [97]
3 years ago
14

Which of the following are trigonometric identities? check all that apply.(Need help asap)

Mathematics
2 answers:
dybincka [34]3 years ago
7 0
A. Yes:

\dfrac{\sin(x+y)}{\sin x\cos y}=\dfrac{\sin x\cos y+\cos x\sin y}{\sin x\cos y}=1+\cot x\tan y

B. Yes:

\dfrac{\csc x-\sin x}{\csc x}=1-\dfrac{\sin x}{\frac1{\sin x}}=1-\sin^2x=\cos^2x

C. Yes:

\tan x\cos x\csc x=\dfrac{\sin x}{\cos x}\cdot\cos x\cdot\dfrac1{\sin x}=1

D. No: if x=0, then 4\cos0\sin0=0, but 2\cos0+1-2\sin0=3.
Ratling [72]3 years ago
4 0

Answer:

(A), (B) and (C)

Step-by-step explanation:

(A) The given statement is:

\frac{sin(x+y)}{sinxcosy}=1+cotxtany

Upon solving, we have

\frac{sin(x+y)}{sinxcosy}=\frac{sinxcosy+cosxsiny}{sinxcosy}=1+cotxtany

Thus, it is a trigonometric identity.

(B) The given statement is:

\frac{cscx-sinx}{cscx}=cos^2x

Upon solving, we have

\frac{cscx-sinx}{cscx}=1-\frac{sinx}{\frac{1}{sinx}}}=1-sin^2x=cos^2x

Thus, it is a trigonometric identity.

(C) The given statement is:

tanxcosxcscx=1

Upon solving, we have

tanxcosxcscx=\frac{sinx}{cosx}{\cdot}cosx{\cdot}\frac{1}{sinx}=1

Thus, it is a trigonometric identity.

(D) The given statement is:

4cosxsinx=2cosx+1-2sinx

Now, if x=0, then

4cos0sin0=0 but 2cos0+1-2sin0=3

Thus, it is not a trigonometric identity.

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