We can also derive the other identities below (1.137) sin−¹(z) = −i ln (iz + √√1 − z²), sinh¯¹(z) = ln (z + √1 + z² - tan−¹(z) = — [ln(1 − iz) − In(1 + iz)] - i tanh¯¹(z) = = [ln(1 + z) — ln(1 − z)]
We can also derive the other identities below (1.137) sin−¹(z) = −i ln (iz + √√1 − z²), sinh¯¹(z) = ln (z + √1 + z² - tan−¹(z) = — [ln(1 − iz) − In(1 + iz)] - i tanh¯¹(z) = = [ln(1 + z) — ln(1 − z)]
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please provide solution for the following equations below.
![We can also derive the other identities below (1.137)
sin−¹(z) = −i ln (iz + √√1 − z²), sinh¯¹(z) = ln (z + √1 + z²
tan−¹(z) = — [ln(1 − iz) − In(1 + iz)]
-
i
tanh¯¹(z) = = [ln(1 + z) — ln(1 − z)]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F244b771d-e8a1-4c98-b5bc-475517a843b9%2F3852adaa-fd2e-4bdf-afdc-c884d9f6aa80%2Fztgopxq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:We can also derive the other identities below (1.137)
sin−¹(z) = −i ln (iz + √√1 − z²), sinh¯¹(z) = ln (z + √1 + z²
tan−¹(z) = — [ln(1 − iz) − In(1 + iz)]
-
i
tanh¯¹(z) = = [ln(1 + z) — ln(1 − z)]
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