SOLUTION: a) prove that (1+tan A + sec A)(1+cot A-cosecA)=2 b) Prove that Sin 20degree.sin40degree.sin 60degree.Sin80degree = 3/16 c) sin 5A+2sin7A+sin9A/Cos3A+2cos5A+Cos7A = Sin2A+Cos2A.T

Algebra ->  Trigonometry-basics -> SOLUTION: a) prove that (1+tan A + sec A)(1+cot A-cosecA)=2 b) Prove that Sin 20degree.sin40degree.sin 60degree.Sin80degree = 3/16 c) sin 5A+2sin7A+sin9A/Cos3A+2cos5A+Cos7A = Sin2A+Cos2A.T      Log On


   

Question 735516: a) prove that (1+tan A + sec A)(1+cot A-cosecA)=2
b) Prove that Sin 20degree.sin40degree.sin 60degree.Sin80degree = 3/16
c) sin 5A+2sin7A+sin9A/Cos3A+2cos5A+Cos7A = Sin2A+Cos2A.Tan5A
d) Show that Sin112+Cos12 / Cos12-Sin12 = Cot35
e) Prove that Sin 7x+Sinx / Cos5x-Cos3x = Sin2x-Cos2x.cotx
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
a) prove that (1+tan A + sec A)(1+cot A-cosecA)=2
LHS =>
+%28%28+cos+A+%2B+sin+A+%2B1%29%2Fcos+A%29%28%28sin+A+%2Bcos+A+-1%29%2Fsin+A%29


+%28%28cos+A+%2Bsin+A+%29%5E2+-1%29%2Fsin+A+cos+A

%28+cos%5E2+%28A+%29%2Bsin+%5E2%28+A%29+%2B2%2Asin+A+cos+A+-1%29%2F%28sin+A+cos+A%29



%281%2B2+sin+A+cosA-1%29%2Fsin+A+cos+A

= 2
LHS = RHS


b) Prove that Sin 20degree.sin40degree.sin 60degree.Sin80degree = 3/16

sin(A+B) = sin A cos B + cos A sin B


sin(40) as sin(60-20)
sin(80) as sin(60+20)

sin(20)•sin(60-20)•[√(3)/2]•sin(60+20)
- Simplify sin(60-20)•sin(60+20)
sin(60-20) = sin60•cos20 - sin20•cos60
sin(60+20)= sin60•cos20 + sin20•cos60
multiply
= sin²60•cos²20 - sin²20•cos²60

cos²A = 1 - sin²A
sin²60•cos²20 - sin²20•cos²60
= sin²60•(1-sin²20) - sin²20•(1-sin²60)
= sin²60- sin²20•sin²60 - sin²20 + sin²20•sin²60
= sin²60 - sin²20
= 3/4 - sin²20

We get
= sin20•[√(3)/2]•(3/4 - sin²20)
= [√(3)/8]•(3sin20 - 4sin³20)

we know that
sin(3A) = 3sin(A) - 4sin³(A)

(3sin20 - 4sin³20) = sin60

we have
= [√(3)/8]•sin60
= [√(3)/8]•[√(3)/2)
= 3/16
LHS = RHS