Question

2√2sin10(sec5/2+cos40/sin5-2sin35)

Answer 1

Let the expression to be evaluated be x.

Since sin10=2sin5cos5, you get:

22–√(sin5+2cos40cos5−4sin35sin5cos5)=22–√(sin5+2cos40cos5−2sin35sin10)

We can write:

2cos40cos5=cos45+cos35

2sin35sin10=cos25−cos45

Hence,

x=22–√(sin5+cos45+cos35+cos45−cos25)=22–√(2–√+sin5+cos35−cos25)

Since

cos35−cos25=−2sin30sin5=−sin5

The final answer is 4

Answer 2

Answer:

Step-by-step explanation: