Use app×
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
0 votes
540 views
in Trigonometry by (46.3k points)
closed by

If 0° < θ < 90° and cos2 θ – sin2 θ = \(\frac12\) , then what is the value of θ?

(a) 30° 

(b) 45° 

(c) 60° 

(d) 90°

1 Answer

+1 vote
by (49.3k points)
selected by
 
Best answer

(a) 30°

cos2 θ – sin2 θ = \(\frac12\)

⇒ (1–sin2 θ) – sin2 θ = \(\frac12\)

⇒ 1– 2 sin2 θ = \(\frac12\)

⇒ 2 sin2 θ = \(\frac12\) ⇒ sin2 θ = \(\frac14\)

⇒ sin θ = \(\frac12\) = sin 30° ⇒ θ = 30°

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

...