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
53 views
in Trigonometry by (71.9k points)
closed by
Find the value of `[(sin^2 2 2^(@)+sin^2 6 8^(@))/(cos^2 2 2^(@)+cos^2 6 8^(@))+sin^2 6 3^(@)+cos6 3^(@)sin2 7^(@)]`
A. 3
B. 2
C. 1
D. 0

1 Answer

0 votes
by (72.1k points)
selected by
 
Best answer
Correct Answer - B
Given, expression , `((sin^(2)22^(@)+sin^(2)68^(@))/(cos^(2)22^(@)+cos^(2)68^(@))+sin^(2)63^(@)+cos63^(@)sin27^(@))`
`(sin^(2)22^(@)+sin^(2)(90^(2)-22^(@)))/(cos^(2)(90^(@)-68^(@))+cos^(2)68^(@))+sin^(2)63^(@)+cos63^(@)sin(90^(@)-63^(@))`
`(sin^(2)22^(@)+cos^(2)22^(@))/(sin^(2)68^(@)+cos^(2)68^(@))+sin^(2)63^(@)+cos63^(@).cos63^(@) [therefore sin(90^(@)-theta)=costheta "and" cos(90^(@)-theta)=sintheta]`
`=1/2+(sin^(2)63^(@)+cos^(2)63^(@))` `[therefore sin^(2)theta+cos^(2)theta=1]`
`1+1=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

...