Question

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

Answer 1

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 $2\cos0+1-2\sin0=3$.

Answer 2

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.