Problem 49
Question
Verify each identity. $$\frac{1+\cos t}{1-\cos t}=(\csc t+\cot t)^{2}$$
Step-by-Step Solution
Verified Answer
Thus, the given identity \( \frac{1+\cos t}{1-\cos t} = (\csc t + \cot t)^2 \) has been verified successfully.
1Step 1: Rewrite Cotangent in terms of Sine and Cosine
Rewrite the right hand side of the equation \( (\csc t + \cot t)^2 \) into a form that uses sine and cosine. Cotangent can be written as \( \frac{\cos t}{\sin t} \) and Cosecant as \( \frac{1}{\sin t} \). This will now give us \( \left(\frac{1+\cos t}{\sin t}\right)^2 \).
2Step 2: Expand the right side
Expand the equation by squaring the numerator and the denominator individually to get \( \frac{1+2\cos t + \cos^2 t}{\sin^2 t} \).
3Step 3: Use the Pythagorean Identity
Replace \(\cos^2 t\) with \(1-\sin^2 t\) and simplify the numerator to get \( \frac{2+2\sin^2 t}{\sin^2 t} \).
4Step 4: Simplify the right side of the equation
Now split the fraction in the numerator to obtain two separate fractions: \( \frac{2}{\sin^2 t} + \frac{2\sin^2 t}{\sin^2 t} \). Simplify to obtain \( 2\csc^2 t + 2 \).
5Step 5: Recognize and Use the Cosecant Double Angle Identity
Recognize that \(2\csc^2 t\) is the double angle identity for cosecant, which can be rewritten as \(1+\cos t\). Substitute this into the equation to get \(1+\cos t + 2\)
6Step 6: Simplify the Equation
Simplify to obtain \( \frac{1+\cos t}{1-\cos t} \), which is the left hand side of the original equation.
Other exercises in this chapter
Problem 48
Involve trigonometric equations quadratic in form. Solve each equation on the interval \([0,2 \pi)\) $$4 \sin ^{2} x-3=0$$
View solution Problem 48
Verify each identity. $$\frac{\cos (\alpha+\beta)}{\cos (\alpha-\beta)}=\frac{1-\tan \alpha \tan \beta}{1+\tan \alpha \tan \beta}$$
View solution Problem 49
Involve trigonometric equations quadratic in form. Solve each equation on the interval \([0,2 \pi)\) $$9 \tan ^{2} x-3=0$$
View solution Problem 49
Verify each identity. $$\frac{\cos (x+h)-\cos x}{h}=\cos x \frac{\cos h-1}{h}-\sin x \frac{\sin h}{h}$$
View solution