Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
51 views
in Trigonometry by (115k points)
closed by
The equation \(\rm sin^{-1}x-cos^{-1}x=\dfrac{\pi}{6}\) has
1. no solution
2. unique solution
3. two solutions
4. infinite number of solutions

1 Answer

0 votes
by (114k points)
selected by
 
Best answer
Correct Answer - Option 2 : unique solution

Concept:

sin-1x + cos-1x = \(\rm \frac{\pi}{2} \)

cos (30∘ ) = \(\rm \frac{\sqrt{3}}{2}\)

Calculation:

Given

⇒ \(\rm sin^{-1}x-cos^{-1}x=\dfrac{\pi}{6}\)

⇒ sin-1x + cos-1x - 2 cos-1x = \(\rm \frac{\pi}{6}\)

⇒ \(\rm \frac{\pi}{2} \)- 2 cos-1x = \(\rm \frac{\pi}{6}\)

⇒ 2 cos-1x  = \(\rm \frac{\pi}{2} - \rm \frac{\pi}{6}\)

⇒ cos-1x = \(\rm \frac{\pi}{6}\)

⇒ x = \(\rm \frac{\sqrt{3}}{2}\)

∴ The equation has unique solution.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...