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
156 views
in Aptitude by (105k points)
closed by
The value of 5 sin 14° sec 76° + 3 cot 15° cot 75° + 2 tan 45° is:

1 Answer

0 votes
by (111k points)
selected by
 
Best answer
Correct Answer - Option 2 : 10

Formulas used:

sec (90 - x) = cosec x

cot (90 - x) = tan x

Calculation:

5 sin 14° sec 76° + 3 cot 15° cot 75° + 2 tan 45°

⇒ 5 sin 14° cosec 14° + cot 15° tan 15° + 2 × 1     [tan 45° = 1]

⇒ 5 sin 14° . 1/sin 14° + 3 cot 15 . 1/cot 15° + 2    [cosec x = 1/sin x]

⇒ 5 + 3 + 2                                                              [cot x = 1/tan x]

∴ The required result = 10

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

...