15. the simplified form of $(\\cos x\\tan x-\\frac{\\sin x}{\\tan x})(\\cos x\\tan x+\\frac{\\sin x}{\\tan…

15. the simplified form of $(\\cos x\\tan x-\\frac{\\sin x}{\\tan x})(\\cos x\\tan x+\\frac{\\sin x}{\\tan x})$ is\na. 1\nc. $\\sin ^{2}x+\\cos ^{2}x$\nb. $1 - 2\\cos ^{2}x$\nd. none of the above\nshort answer\n16. write the expression $\\tan x\\sin x$ in terms of $\\cos x$.\n17. write the expression $\\frac{\\tan ^{2}x}{1+\\tan ^{2}x}$ in terms of $\\sin x$.\n18. simplify $\\frac{\\cos x}{1 - \\sin x}-\\tan x$.\n19. simplify $\\frac{1}{\\sin ^{2}x}-\\frac{1}{\\tan ^{2}x}$.\n20. simplify $\\frac{\\cos x}{\\tan x}+\\sin x$.\n21. simplify $1+\\frac{1}{\\tan ^{2}x}$.\n22. simplify $\\frac{1-\\frac{1}{\\cos ^{2}x}}{1 - \\cos ^{2}x}$.\n23. prove that $\\tan x+\\frac{1}{\\tan x}=\\frac{1}{\\sin x\\cos x}$.

15. the simplified form of $(\\cos x\\tan x-\\frac{\\sin x}{\\tan x})(\\cos x\\tan x+\\frac{\\sin x}{\\tan x})$ is\na. 1\nc. $\\sin ^{2}x+\\cos ^{2}x$\nb. $1 - 2\\cos ^{2}x$\nd. none of the above\nshort answer\n16. write the expression $\\tan x\\sin x$ in terms of $\\cos x$.\n17. write the expression $\\frac{\\tan ^{2}x}{1+\\tan ^{2}x}$ in terms of $\\sin x$.\n18. simplify $\\frac{\\cos x}{1 - \\sin x}-\\tan x$.\n19. simplify $\\frac{1}{\\sin ^{2}x}-\\frac{1}{\\tan ^{2}x}$.\n20. simplify $\\frac{\\cos x}{\\tan x}+\\sin x$.\n21. simplify $1+\\frac{1}{\\tan ^{2}x}$.\n22. simplify $\\frac{1-\\frac{1}{\\cos ^{2}x}}{1 - \\cos ^{2}x}$.\n23. prove that $\\tan x+\\frac{1}{\\tan x}=\\frac{1}{\\sin x\\cos x}$.

Answer

Problem 15

Explanation:

Step1: Use the difference of squares formula ((a - b)(a + b)=a^{2}-b^{2})

Let (a = \cos x\tan x) and (b=\frac{\sin x}{\tan x}). Then (\left(\cos x\tan x-\frac{\sin x}{\tan x}\right)\left(\cos x\tan x+\frac{\sin x}{\tan x}\right)=(\cos x\tan x)^{2}-\left(\frac{\sin x}{\tan x}\right)^{2})

Step2: Simplify (\cos x\tan x) and (\frac{\sin x}{\tan x})

Since (\tan x=\frac{\sin x}{\cos x}), (\cos x\tan x=\cos x\times\frac{\sin x}{\cos x}=\sin x), and (\frac{\sin x}{\tan x}=\frac{\sin x}{\frac{\sin x}{\cos x}}=\cos x)

Step3: Substitute back into the expression

((\cos x\tan x)^{2}-\left(\frac{\sin x}{\tan x}\right)^{2}=\sin^{2}x-\cos^{2}x) Since (\sin^{2}x+\cos^{2}x = 1), (\sin^{2}x-\cos^{2}x=1 - 2\cos^{2}x)

Answer:

b. (1 - 2\cos^{2}x)

Problem 16

Explanation:

Step1: Express (\tan x) in terms of (\sin x) and (\cos x)

Since (\tan x=\frac{\sin x}{\cos x}), then (\tan x\sin x=\frac{\sin x}{\cos x}\times\sin x)

Answer:

(\frac{\sin^{2}x}{\cos x})

Problem 17

Explanation:

Step1: Use the identity (1+\tan^{2}x=\sec^{2}x=\frac{1}{\cos^{2}x})

(\frac{\tan^{2}x}{1 + \tan^{2}x}=\tan^{2}x\cos^{2}x)

Step2: Express (\tan x) in terms of (\sin x) and (\cos x)

Since (\tan x=\frac{\sin x}{\cos x}), (\tan^{2}x\cos^{2}x=\left(\frac{\sin x}{\cos x}\right)^{2}\cos^{2}x)

Answer:

(\sin^{2}x)

Problem 18

Explanation:

Step1: Express (\tan x) as (\frac{\sin x}{\cos x})

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

Step2: Find a common denominator

The common denominator is (\cos x(1-\sin x)). So (\frac{\cos^{2}x-\sin x(1 - \sin x)}{\cos x(1-\sin x)}=\frac{\cos^{2}x-\sin x+\sin^{2}x}{\cos x(1-\sin x)})

Step3: Use the identity (\sin^{2}x+\cos^{2}x = 1)

(\frac{1-\sin x}{\cos x(1-\sin x)})

Answer:

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

Problem 19

Explanation:

Step1: Express (\tan x) as (\frac{\sin x}{\cos x})

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

Answer:

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

Problem 20

Explanation:

Step1: Express (\tan x) as (\frac{\sin x}{\cos x})

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

Step2: Simplify the first - term

(\frac{\cos^{2}x}{\sin x}+\sin x=\frac{\cos^{2}x+\sin^{2}x}{\sin x})

Step3: Use the identity (\sin^{2}x+\cos^{2}x = 1)

Answer:

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

Problem 21

Explanation:

Step1: Express (\tan x) as (\frac{\sin x}{\cos x})

(1+\frac{1}{\tan^{2}x}=1+\frac{\cos^{2}x}{\sin^{2}x})

Step2: Find a common denominator

(\frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x})

Step3: Use the identity (\sin^{2}x+\cos^{2}x = 1)

Answer:

(\frac{1}{\sin^{2}x}=\csc^{2}x)

Problem 22

Explanation:

Step1: Simplify the numerator and denominator

The numerator (1-\frac{1}{\cos^{2}x}=\frac{\cos^{2}x - 1}{\cos^{2}x}), and the denominator (1-\cos^{2}x=\sin^{2}x)

Step2: Substitute and simplify

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

Answer:

(-\sec^{2}x)

Problem 23

Explanation:

Step1: Simplify the left - hand side

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

Step2: Find a common denominator

The common denominator is (\sin x\cos x). So (\frac{\sin^{2}x+\cos^{2}x}{\sin x\cos x})

Step3: Use the identity (\sin^{2}x+\cos^{2}x = 1)

Answer:

Since (\frac{\sin^{2}x+\cos^{2}x}{\sin x\cos x}=\frac{1}{\sin x\cos x}), the identity (\tan x+\frac{1}{\tan x}=\frac{1}{\sin x\cos x}) is proved.