plz help to solve
how to proceed further plz say guys
Answers
Answer:
cos20.cos40.cos80 / 1
Multiplying Both Numerator And Denominator by 2Sin20
cos20.cos40.cos80⇒(2Sin20.cos20).cos40.cos80
---------------------------------------
2Sin20
⇒sin(2 * 20).cos40.cos80 [2SinA.CosA = Sin 2A]
-------------------------------- ↓
2sin20
⇒sin40.cos40.cos80 You Would Know This
-------------------------- If u Are In 11th or 12th
2sin20
[Multiplying both the numerator and denominator by 2]
⇒(2Sin40.Cos40).cos80
----------------------------- [2SinA.CosA = Sin 2A]
2 * 2sin20
⇒Sin(2 * 40).cos80
----------------------
4Sin20
⇒ Sin80.cos80
------------------
4Sin20
[Multiplying both the numerator and denominator by 2]
⇒2*Sin80.cos80
--------------------- [2SinA.CosA = Sin 2A]
2 * 4Sin20
⇒Sin(2 * 80)
--------------
8 Sin 20
⇒Sin 160
-----------
8 Sin20
⇒Sin (180 -20)
----------------- [Sin(180 - A) = Sin A]
8 SIn 20
⇒Sin 20 / 8 Sin 20
⇒1/8
Hope U Understood Plz Any Doubt About This Then Plz Message Me And Mark It Brainliest if this was helpful
Answer:
♥️♠️ HELLO JII ♥️♠️
cos20 cos40 cos80
multiply and divide by 2
1/2 (2 cos20 cos40 cos80)
1/2 (cos(20+80)+ cos(20-80)) cos40 )
(2cosa cosb= cos(a+b) + cos(a-b) FORMULA
1/2 (cos(-60)+ cos(100)) cos40
1/2 (1/2 + cos100)cos40
1/4 cos40+ 1/2 (cos40 cos100)
multiplt and divide by 2
2/2(1/4 cos40) + 1/4 (2 cos40 cos100)
1/4 cos40+ 1/4 (cos140+ cos(-60))
(2cosa cosb= cos(a+b) cos(a-b)) FORMULA
1/4 cos40+ 1/4 cos140 + 1/8 (cos60= 1/2)
1/4 (cos40+cos140) + 1/8
1/4(2 cos90 cos(-50)) + 1/8 (as above identity)
cos90= 0
= 1/8 r.h.s
hence proved
Step-by-step explanation: