Question

The expressions A, B, C, D, and E are left-hand sides of trigonometric identities. The expressions 1, 2, 3, 4, and 5 are right-hand side of identities. Match each of the left-hand sides below with the appropriate right-hand side.
A. tan(x)
B. cos(x)
C. sec(x)csc(x)
D. 1â(cos(x))^2/ cos(x)
E. 2sec(x)

1. sin(x)tan(x)
2. sin(x)sec(x)
3. tan(x)+cot(x)
4. cos(x)/1âsin(x)+1âsin(x)/cos(x)
5. sec(x)âsec(x)(sin(x))2

Answer 1

Answer:

$A.\ \tan(x) \to 2.\ \sin(x) \sec(x)$

$B.\ \cos(x) \to 5. \sec(x) - \sec(x)\sin^2(x)$

$C.\ \sec(x)csc(x) \to 3. \tan(x) + \cot(x)$

$D. \frac{1 - (cos(x))^2}{cos(x)} \to 1. \sin(x) \tan(x)$

$E.\ 2\sec(x) \to\ 4.\ \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}$

Step-by-step explanation:

Given

$A.\ \tan(x)$

$B.\ \cos(x)$

$C.\ \sec(x)csc(x)$

$D.\ \frac{1 - (cos(x))^2}{cos(x)}$

$E.\ 2\sec(x)$

Required

Match the above with the appropriate identity from

$1.\ \sin(x) \tan(x)$

$2.\ \sin(x) \sec(x)$

$3.\ \tan(x) + \cot(x)$

$4.\ \frac{cos(x)}{1 - sin(x)} + \frac{1 - \sin(x)}{cos(x)}$

$5.\ \sec(x) - \sec(x)(\sin(x))^2$

Solving (A):

$A.\ \tan(x)$

In trigonometry,

$\frac{sin(x)}{\cos(x)} = \tan(x)$

So, we have:

$\tan(x) = \frac{\sin(x)}{\cos(x)}$

Split

$\tan(x) = \sin(x) * \frac{1}{\cos(x)}$

In trigonometry

$\frac{1}{\cos(x)} =sec(x)$

So, we have:

$\tan(x) = \sin(x) * \sec(x)$

$\tan(x) = \sin(x) \sec(x)$ --- proved

Solving (b):

$B.\ \cos(x)$

Multiply by $\frac{\cos(x)}{\cos(x)}$ --- an equivalent of 1

So, we have:

$\cos(x) = \cos(x) * \frac{\cos(x)}{\cos(x)}$

$\cos(x) = \frac{\cos^2(x)}{\cos(x)}$

In trigonometry:

$\cos^2(x) = 1 - \sin^2(x)$

So, we have:

$\cos(x) = \frac{1 - \sin^2(x)}{\cos(x)}$

Split

$\cos(x) = \frac{1}{\cos(x)} - \frac{\sin^2(x)}{\cos(x)}$

Rewrite as:

$\cos(x) = \frac{1}{\cos(x)} - \frac{1}{\cos(x)}*\sin^2(x)$

Express $\frac{1}{\cos(x)}\ as\ \sec(x)$

$\cos(x) = \sec(x) - \sec(x) * \sin^2(x)$

$\cos(x) = \sec(x) - \sec(x)\sin^2(x)$ --- proved

Solving (C):

$C.\ \sec(x)csc(x)$

In trigonometry

$\sec(x)= \frac{1}{\cos(x)}$

and

$\csc(x)= \frac{1}{\sin(x)}$

So, we have:

$\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)}$

Multiply by $\frac{\cos(x)}{\cos(x)}$ --- an equivalent of 1

$\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)} * \frac{\cos(x)}{\cos(x)}$

$\sec(x)csc(x) = \frac{1}{\cos^2(x)}*\frac{\cos(x)}{\sin(x)}$

Express $\frac{1}{\cos^2(x)}\ as\ \sec^2(x)$ and $\frac{\cos(x)}{\sin(x)}\ as\ \frac{1}{\tan(x)}$

$\sec(x)csc(x) = \sec^2(x)*\frac{1}{\tan(x)}$

$\sec(x)csc(x) = \frac{\sec^2(x)}{\tan(x)}$

In trigonometry:

$tan^2(x) + 1 =\sec^2(x)$

So, we have:

$\sec(x)csc(x) = \frac{\tan^2(x) + 1}{\tan(x)}$

Split

$\sec(x)csc(x) = \frac{\tan^2(x)}{\tan(x)} + \frac{1}{\tan(x)}$

Simplify

$\sec(x)csc(x) = \tan(x) + \cot(x)$  proved

Solving (D)

$D.\ \frac{1 - (cos(x))^2}{cos(x)}$

Open bracket

$\frac{1 - (cos(x))^2}{cos(x)} = \frac{1 - cos^2(x)}{cos(x)}$

$1 - \cos^2(x) = \sin^2(x)$

So, we have:

$\frac{1 - (cos(x))^2}{cos(x)} = \frac{sin^2(x)}{cos(x)}$

Split

$\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \frac{sin(x)}{cos(x)}$

$\frac{sin(x)}{\cos(x)} = \tan(x)$

So, we have:

$\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \tan(x)$

$\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) \tan(x)$ --- proved

Solving (E):

$E.\ 2\sec(x)$

In trigonometry

$\sec(x)= \frac{1}{\cos(x)}$

So, we have:

$2\sec(x) = 2 * \frac{1}{\cos(x)}$

$2\sec(x) = \frac{2}{\cos(x)}$

Multiply by $\frac{1 - \sin(x)}{1 - \sin(x)}$ --- an equivalent of 1

$2\sec(x) = \frac{2}{\cos(x)} * \frac{1 - \sin(x)}{1 - \sin(x)}$

$2\sec(x) = \frac{2(1 - \sin(x))}{(1 - \sin(x))\cos(x)}$

Open bracket

$2\sec(x) = \frac{2 - 2\sin(x)}{(1 - \sin(x))\cos(x)}$

Express 2 as 1 + 1

$2\sec(x) = \frac{1+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}$

Express 1 as $\sin^2(x) + \cos^2(x)$

$2\sec(x) = \frac{\sin^2(x) + \cos^2(x)+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}$

Rewrite as:

$2\sec(x) = \frac{\cos^2(x)+1 - 2\sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}$

Expand

$2\sec(x) = \frac{\cos^2(x)+1 - \sin(x)- \sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}$

Factorize

$2\sec(x) = \frac{\cos^2(x)+1(1 - \sin(x))- \sin(x)(1-\sin(x))}{(1 - \sin(x))\cos(x)}$

Factor out 1 - sin(x)

$2\sec(x) = \frac{\cos^2(x)+(1- \sin(x))(1-\sin(x))}{(1 - \sin(x))\cos(x)}$

Express as squares

$2\sec(x) = \frac{\cos^2(x)+(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}$

Split

$2\sec(x) = \frac{\cos^2(x)}{(1 - \sin(x))\cos(x)} +\frac{(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}$

Cancel out like factors

$2\sec(x) = \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}$ --- proved