0 votes
121 views
in Chapter 2 Inverse Trigonometric Functions by (8.1k points)
edited

If sin(sin-1\(\frac { 1 }{ 5 }\) + cos-1x) = 1, then find the value of x.

1 Answer

0 votes
by (8.1k points)
selected by
 
Best answer

sin(sin-1\(\frac { 1 }{ 5 }\) + cos-1x) = sin\(\frac { π }{ 2 }\)
⇒ sin-1\(\frac { 1 }{ 5 }\) + cos-1x = sin\(\frac { π }{ 2 }\)
or (sin-1\(\frac { 1 }{ 5 }\) + cos-1\(\frac { 1 }{ 5 }\)) + (- cos-1\(\frac { 1 }{ 5 }\)x) = \(\frac { π }{ 2 }\)
or \(\frac { π }{ 2 }\) – cos-1\(\frac { 1 }{ 5 }\) + cos-1x = \(\frac { π }{ 2 }\)
or cos-1\(\frac { 1 }{ 5 }\) – cos-1x = \(\frac { π }{ 2 }\)
or cos-1x = cos-1\(\frac { 1 }{ 5 }\)
⇒ x = \(\frac { 1 }{ 5 }\)

Related questions

Doubtly is an online community for engineering students, offering:

  • Free viva questions PDFs
  • Previous year question papers (PYQs)
  • Academic doubt solutions
  • Expert-guided solutions

Get the pro version for free by logging in!

5.7k questions

5.1k answers

108 comments

561 users

...