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show that tan70-tan20=2tan40+4tan10 please work out with this quetion

show that tan70-tan20=2tan40+4tan10 please work out with this quetion
                                                                                                                                                 

Grade:11

3 Answers

somi teez
105 Points
6 years ago
hi,
tan70−tan20−2tan40=(tan70−tan40)−(tan40+tan20)=sin70cos70−sin40cos40−(sin40cos40+sin20cos20)=sin30cos70cos40−sin60cos20cos40=12cos40(1cos70−3–√cos20)=12cos40(cos20−3–√cos70cos70cos20)=12cos40(cos20−3–√sin20sin20cos20)=1cos40⎛⎝12cos20−3√2sin20sin20cos20⎞⎠=2cos40cos60cos20−sin60sin202sin20cos20=2cos40cos80sin40=4cos802sin40cos40=4cos80sin80=4cot80=4tan10tan70−tan20−2tan40=(tan70−tan40)−(tan40+tan20)=sin70cos70−sin40cos40−(sin40cos40+sin20cos20)=sin30cos70cos40−sin60cos20cos40=12cos40(1cos70−3–√cos20)=12cos40(cos20−3–√cos70cos70cos20)=12cos40(cos20−3–√sin20sin20cos20)=1cos40⎛⎝12cos20−3√2sin20sin20cos20⎞⎠=2cos40cos60cos20−sin60sin202sin20cos20=2cos40cos80sin40=4cos802sin40cos40=4cos80sin80=4cot80=4tan10
somi teez
105 Points
6 years ago
its better to write tan 70 -tan 20 -2 tan 40 = 4 tan 10 (to be proved)
now,
lhs,
=(tan 70 – tan 40 ) – (tan 40 + tan 20)
=(sin 70 /cos 70 – sin 40/ cos 40 ) – (sin 40 / cos 40 + sin 20 / cos 20)
=sin 30 /(cos 70 *cos 40)  – sin 60 / (cos 20 * cos 40 )
=1/ 2 cos 40 *(1/cos 70 – sqrt 3 /cos 20)
=1/cos 40 *(1/2*cos 20 -sqrt 3*sn 20 /2)/(sin 20* cos 20)
=2/cos 40 *(cos 60* cos 20 -sin 60 * sin 20 )/(2 sin 20 *cos 20)
=2/ cos 40 *(cos 80/ sin 40)
=4 *(cos 80/ 2 cos 40* sin 40)
=4*(cos 80 /sin 80)
=4 cot 80
=4 cot (90-10
=4 tan 10 …........hence proved
i used basic addition and subtraction of trigo... please understand  ..as i skipped some steps
approve it if you agree
BEST OF LUCK!!
Akash Singh
15 Points
3 years ago
Sol. tan70-tan20-2tan40=4tan10
 LHS, tan70=cot20=1/tan20
 = (1/tan20 -tan20)-2tan40
 =  {(1-tan220)/tan20}-2tan40
 = {2/(2tan20/1-tan220)-2tan40  (multi.÷ by2
  apply formula of tan2x=2tanx/1-tan2x
 = {2/tan40}-2tan40
 Again apply same procedure
 = 2{(1-tan240)/tan40}
 = 2{1×2/(2×tan40/1-tan240)}
 =2{2/tan80} (apply tan2x formula)
 =4cot80=4cot(90-10)
 =4tan10=RHS
Hence proved
It's easy one just convert into tan2x formula
& apply it twice....

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