Register

Help me please ( \cos(80) )/( \sin(10) ) + ( \sin(20) )/( \cos(70) ) = 2 \\ \\ ( \cot(40) )/( \tan(50) ) + ( \cos(65) )/( \sin(115) ) = 2 ​

Help me please ( \cos(80) )/( \sin(10) ) + ( \sin(20) )/( \cos(70) ) = 2 \\ \\ ( \cot(40) )/( \tan(50) ) + ( \cos(65) )/( \sin(115) ) = 2 ​
asked 166k views
4 votes
Help me please


( \cos(80) )/( \sin(10) ) + ( \sin(20) )/( \cos(70) ) = 2 \\ \\ ( \cot(40) )/( \tan(50) ) + ( \cos(65) )/( \sin(115) ) = 2
​

asked
User Oscarkuo
by
7.3k points

1 Answer

1 vote

Answer:


( \cos(80) )/( \sin(10) ) + ( \sin(20) )/( \cos(70) ) = 2 \\ ( \cos(90 - 10) )/( \sin(10) ) + ( \sin(20) )/( \cos(9 0 - 20) ) = 2 \\ ( \sin(10) )/( \sin(10) ) + ( \sin(20) )/( \sin(20) ) = 2 \\ 1 + 1 = 2 \\ 2 = 2 \\ \\ ( \cot(40) )/( \tan(50) ) + ( \cos(65) )/( \sin(115) ) = 2 \\ ( \cot(90 - 50) )/( \tan(50) ) + ( \cos(65) )/( \sin(90 + 65) ) = 2 \\ ( \tan(50) )/( \tan(50) ) + ( \cos(65) )/( \cos(65) ) = 2 \\ 1 + 1 \\ = 2

answered
User Ryder Mackay
by
7.7k points
← Prev Question Next Question →

No related questions found

Ask a Question
Welcome to Qamnty — a place to ask, share, and grow together. Join our community and get real answers from real people.

Categories

Other Questions

How do you can you solve this problem 37 + y = 87; y =
How do you estimate of 4 5/8 X 1/3
A bathtub is being filled with water. After 3 minutes 4/5 of the tub is full. Assuming the rate is constant, how much longer will it take to fill the tub?
Link Copied!