Use app×
Ask a Question
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE

Evaluate: (cosec^2 63° + tan^2 24°) / (cot^2 66° + sec^2 27°) + (sin^2 63° + cos 63° sin 27° + sin 27° sec 63°)

← Prev Question Next Question →
+1 vote
24.5k views
in Mathematics by (83.5k points)

Evaluate: 

(cosec2 63° + tan2 24°) / (cot2 66° + sec2 27°) + (sin2 63° + cos 63° sin 27° + sin 27° sec 63°) / (2(cosec2 65° - tan2 25°))

Challenge Your Friends with Exciting Quiz Games – Click to Play Now!

1 Answer

+1 vote
by (80.0k points)
selected by
 
Best answer

(cosec2 63° + tan2 24°) / (cot2 66° + sec2 27°) + (sin2 63° + cos 63° sin 27° + sin 27° sec 63°) / (2(cosec2 65° - tan2 25°))

← Prev Question Next Question →

Find MCQs & Mock Test

  1. JEE Main 2026 Test Series
  2. NEET Test Series
  3. Class 12 Chapterwise MCQ Test
  4. Class 11 Chapterwise Practice Test
  5. Class 10 Chapterwise MCQ Test
  6. Class 9 Chapterwise MCQ Test
  7. Class 8 Chapterwise MCQ Test
  8. Class 7 Chapterwise MCQ Test

Related questions

0 votes
1 answer
Prove that tan^2 66° − cot^2 24° = 0.
asked Apr 19, 2020 in Trigonometry by Vevek01 (44.3k points)

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

User Contribution to Sarthaks eConnect
Managed by Sarthaks Digitizing India
...