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
87 views
in Aptitude by (114k points)
closed by

If (secθ + sinθ)/(secθ - sinθ) = (4 + √3)/(4 - √3), 0°<θ < 90°, then, what is the value of (cos2θ + √3tanθ)?


1. 3/2
2. 2/3
3. 4/5
4. 4/3

1 Answer

0 votes
by (113k points)
selected by
 
Best answer
Correct Answer - Option 1 : 3/2

Given:

(secθ + sinθ)/(secθ - sinθ) = (4 + √3)/(4 - √3), 0°<θ < 90°

Formula Used:

(a +b)/(a - b) = c/d

∴ a/b = (c + d)/(c – d)

Calculation:

(secθ + sinθ)/(secθ - sinθ) = (4 + √3)/(4 - √3)

By componendo and dividendo

⇒ (secθ + sinθ + secθ - sinθ)/(secθ + sinθ - secθ + sinθ) = (4 + √3 + 4 - √3)/(4 + √3 - 4 + √3)

⇒ 2secθ/2sinθ = (2 × 4)/(2 × √3)

secθ/sinθ = 4/√3

⇒ sinθ/secθ = √3/4

⇒ sinθ cosθ = √3/4

⇒ 2sinθ cosθ = √3/2

⇒ sin2θ = √3/2

⇒ 2θ = 60°

⇒ θ = 30°

The value of (cos2θ + √3tanθ) = cos60° + √3tan30°

⇒ (1/2) + √3(1/√3)

⇒ 1/2 + 1

⇒ 3/2

∴ The correct answer is 3/2  

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

...