Question

show that tan 70° - tan 20° = 2tan 40° + 4tan 10°.​

Answer 1

Answer:

4 tan 10

Step-by-step explanation:

= (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

Answer 2

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....