Question
Prove that each equation is an identity.$$\sec (\alpha+\beta)=\frac{\sec \alpha \sec \beta}{1-\tan \alpha \tan \beta}$$
Step 1
Step 1: We start with the right-hand side of the equation and convert the secants to their equivalent in terms of cosine: $$\frac{\sec \alpha \sec \beta}{1-\tan \alpha \tan \beta} = \frac{\frac{1}{\cos \alpha} \cdot \frac{1}{\cos \beta}}{1-\frac{\sin \alpha}{\cos Show more…
Show all steps
Your feedback will help us improve your experience
Subhadeepta Sahoo and 79 other Precalculus educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Verify that each equation is an identity. $\frac{\tan (\alpha+\beta)-\tan \beta}{1+\tan (\alpha+\beta) \tan \beta}=\tan \alpha$
Trigonometric Identities
Sum and Difference Identities for Sine and Tangent
Verify that each equation is an identity. $$\frac{\tan (\alpha+\beta)-\tan \beta}{1+\tan (\alpha+\beta) \tan \beta}=\tan \alpha$$
Verify that each equation is an identity. $$\frac{\cos (\alpha+\beta)}{\sin (\alpha-\beta)}=\frac{1-\tan \alpha \tan \beta}{\tan \alpha-\tan \beta}$$
Trigonometric ldentities and Conditional Equations
Sum and Difference Identities
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD